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AN ELEMENTARY TREATISE ON THE
GEOMETRY OF CONICS.
AN ELEMENTARY TREATISE
ON THE
GEOMETRY OF CONICS.
BY
ASUTOSH ArcKHOPADHYAY, ALA., F.RS.K,
PIIEMCHAXD BOTCHAXD 3TUDEXT, rELLOW", AND MEMBER OK 1 HE ST^IDICATE OP THE
OTVERSirr OF Calcutta., pkllow of the royal astboxomical socibtv,
MEMBER OF THE ROYAL IRISH ACADEMY, OF THE MATHEMATICAL
SOCIKTT OF FRASCE, ETC, ETC.
#
MACMILLAX AND CO..
AND NEW YORK.
1893.
All rights reserved.
PREFACE.
This work contains elementary proofs of the principal
properties of Conies, and is intended for students who
proceed to the study of the subject after finishing the
first six books of Euclid : the curves have not, therefore,
been defined as the sections of a cone, although that
method has the sanction of history and antiquity in its
favour ; and for the same reason, no use has been made
of the method of projections.
As regards the arrangement of the subject, I have
thought it best to devote separate chapters to the para-
bola, the ellipse, and the h}^erbola. The plan of starting
with a chapter on general conies, in which some funda-
mental propositions are proved by methods applicable
to all the three curves, has no doubt the advantage of
securing an appearance of brevity. But, I believe, be-
ginners find the subject more iatelligible when the pro-
perties of the three cur\'e.s are discussed separately.
Besides, in the other method students, and even writers
of text-books, are apt to overlook the necessity of
modifying an argument on account of the fundamental
vi GEOMETRY OF CONICS.
difference in the figures of the several curves ; see, for
instance, Chap. II,, Prop, x., and Chap. III., Prop, ix.,
which are ordinarily proved by identically the same
argument. Also, as the properties of the hyperbola are
proved, wherever possible, by the same methods as the
corresponding properties of the ellipse, it is obvious that
this arrangement does not tend to increase the work of
the student.
As to the propositions included in each chapter and
their sequence, I have not been able to adopt wholly the
scheme of any previous writer; but I venture to hope
that the book includes all the classical propositions on
the subject, arranged in their proper logical order.
Every attempt has been made to render the proofs
simple and easily intelligible, though I have never sacri-
ficed accuracy to brevity. Thus, for instance, I have
not followed the practice of referring to a proposition
when the truth of its converse is really assumed — a
practice which has, in at least one instance, led to a
remarkable error in the treatment of conjugate dia-
meters in a famous text-book. Nor have I attempted
to secure a fictitious appearance of conciseness by adding
to each proposition a list of corollaries by no means less
important than the proposition itself, and freely using
them for the purpose of deducing subsequent proposi-
tions.
The exercises, of which there are about eight hundred,
have been selected with great care ; more than six hun-
dred of these are placed under the different propositions
from which they may be deduced ; they are for the most
PREFACE.
Vll
part of an elementary character, and have been carefully
graduated. Hints and solutions have been liberally
added, and these, it is hoped, will prove materially
helpful to the student, and render the subject attractive.
The attention of the student has also been directed to
various methods of graphically describing the curves,
including those used in practice by draughtsmen, and
some very neat problems have been added from Newton,
Book I., Sections iv. and v.
At the end of the table of contents will be found a
course of reading suitable for beginners.
Calcutta,
19/A April, 1893.
CONTENTS.
ISTRODUCTIOK,
FAUB
1
CHAPTER I.
THE PARABOLA.
Description of the Curve,
3
Properties of Chords, ......
6
Properties of Tangents,
20
Properties of Normals, ......
44
Miscellaneous Examples
47
CHAPTER n.
THE K L L I P S K.
Description of the Curve, .50
Properties of Chords and their Segments, .... 54
Pi'operties of Tangents, 70
Properties of Xomials, ... .... 90
Properties of Conjugate Diameters, 94
Miscellaneous Examples, . I(i4
CHAPTER III.
THE HYPERBOLA.
Description of the Curve,
Properties of Chords and their Segments,
Properties of Tangents. ....
Properties of Normals, ....
110
115
129
147
GEOMETRY OF CONICS.
Properties of Asymptotes, 1^2
Properties of Conjugate Diameters, 163
The Equilateral Hyperbola, I'l
Miscellaneous Examples, ^'^
Propositions marked with an asterisk may be omitted by the
beginner. Tliis would leave for a first course of reading —
Chap. I.— Props, i.-ii., iv.-vii., x.-xii., xiv., xvii.-xix., xxiii.-
XXV., (16)
Chap. II. — Props, i.-v., viii.-xi., xiv.-xix.. xxi.-xxiii., xxv.,
xxvi., XXX., xxxi., xxxiii., xxxiv., . . (24)
Chap. III.— Props, i.-iv., vii.-ix., xii.-xvii., xix.-xxi., xxiii.,
xxvii.-xxxi., xxxiii. -xxxvi, A — !>., . . • (30)
GEOMETRY OF CONICS.
GEOMETRY OF CONICS.
INTRODUCTION.
A (jONic is a curve traced by a point which moves in a
plane containing a fixed point and a fixed straight line,
in such a way that its distance from the fixed point is in
a constant ratio to its perpendicular distance from the
fixed straight line.
The fixed point is called the Focus.
The fixed straight line is called the Dieectrix.
The constant ratio is called the Eccentbicity, and is
usually represented by the letter e.
When the eccentricity is equal to unity, the Conic is
called a Parabola (e=l).
When the eccentricity is less than unity, the Conic is
called an Ellipse (e < 1).
When the eccentricity is greater than unity, the Conic
is called a Hyperbola (^ > 1).
The straight line drawn through the focus perpen-
dicular to the directrix is called the Axis of the Conic.
The point (or points) in which the axis intersects the
Conic is called the Vertex.
The Conies are so called from the circumstance that thej are,
and were originaUj studied as, the plane sections of the surface
A
2 GEOMETRY OF CONICS.
of a right circular cone, which is a surface formed by the revol-
ution of a right-angled triangle about one of its sides. This
conception does not lead to the simplest way of investigating
the properties of Conies, as it necessitates a knowledge of the
geometry of solids. In order to restrict the discussion of these
curves to the domain of plane geometry, they have been defined
as above.
The Conies are said to have been discovered by Menaechmus,
a Greek mathematician who flourished about B.C. 350, and were
accordingly called after him the " Menaechmian Triads.^' They
were first systematically studied by Apollonius of Perga (b.c.
247-206).
CHAPTER I.
THE PARABOLA.
DESCEIPTION OF THE CURVE.
We have seen that the eccentricity of the parabola is
unity, that is, the distance of any point on it from the
focus is equal to its perpendicular distance from the
directrix.
The parabola may be mechanically constructed in the
following manner.
Let /S be the focus and MX the directrix ; and let a
rigid bar KMQ, of which the portions KM and MQ are
at right angles to each other, having a string of the same
length as MQ, fastened at the end Q, be made to slide
3
4 GEOMETEY OF CONICS.
parallel to the axis SX with the end M on the directrix ;
then if the other end of the string be fastened at the
focus S, and the string be kept stretched by means of
the point of a pencil at P, in contact with the bar, it
is evident that the point P will trace out a parabola,
since SP is always equal to PM.
Ex. A point moves so that the sum of its distances from a
fixed point and a fixed straight line is constant. Show that it
describes a parabola.
In the above figure, the siun of the distances of P from S and
the straight line tlirough Q parallel to XK is evidently constant.
Proposition I.
Given the focus and the directrix of a parabola, to
determine any nuwher of points on it.
Let S be the focus and MXM' the directrix. Through
S draw ^X perpendicular to the directrix, and bisect SX
in A ; then ^ is a point on the parabola, since 8A=AX.
Take any point N in SX or SX produced. Through
F draw PNP' perpendicular to XN ) with centre /Si and
PAEABOLA. 5
radins equal to XN, describe a circle cutting PNP' at P
and P' ; then P and P' shall be points on the parabola.
Draw PM and P'lT perpendicular to the directrix.
Then PS^JlIv', by construction, and PM=XN, being
opposite sides of a rectangle ; therefore PS=PJI. Simi-
larly it may be shown that FS=P'M'. Therefore P and
P' are points on the parabola.
In like manner, by taking any other point in SX, any
number of points on the curve may be determined.
Ex. 1. The parabola is svmmetrical with respect to its axis.
This follows from the fact that PF is bisected at right angles
De£ A curve ls said to be symmetrical luith respect
to a straight line, if, corresponding to any point on the
curve, there is another point on the curve on the other
side of the straight line, such that the chord joining them
is bisected at right angles by the straight lin&
Ex. 2. Alternative Cojutructum — Join the f ocas S to any point
Mon the directrix ; draw JfP at right angles to the directrix, and
make the angle MSP equal to the angle SMP. P is a point on
the parabola.
Ex. 3. Alternative Construction. — Bisect SM in E, and draw
EP perpendicidar to SM, meeting MP in P. P is a point on the
parabola.
For another construction, see Prop. X., Ex. 3.
Ex. 4. Describe a parabola of which the focus and vertex are
given.
Ex. 5. Given the focus S, and two points P, Q on the parabola,
construct it.
The directrix will be a common tangent to the two circles
described, with centres P, Q and radii SP, SQ respectively.
Ex. 6. The distance of any point inside the parabola from the
focus is less than its distance from the directrix.
Ex. 7. The distance of any point outside the parabola from the
focus is greater than its distance from the directrix.
Ex. 8. A straight line parallel to the axis of a parabola meets
the curve in one point only.
Ex. 9. There is no limit to the distance to which the parabola
e GEOMETRY OF CONICS.
may extend on both sides of the axis, so that the parabola is not
a dosed curve.
It is obvious that the point If may be taken ani/where on the
axis.
Ex. 10. Any two right lines drawn from the focus to the curve
on opposite sides of the axis, and equally inclined to it, are equal ;
and conversely.
Ex. 11. If S3f meets in T the straight line drawn through A
perpendicular to the axis, SY=FJI, and FY is at right angles to
SM and bisects the angle SPM.
Ex. 12. If SZ is drawn at right angles to SP to meet the
directrix in Z, PZ bisects the angle SPM.
Ex. 13. PSp is a right line passing through the focus and
meeting the parabola in P and 'p. PM and pm are perpendicular
to the directrix. Show that MHm is a right angle.
Ex. 14. The locus of the centre of a circle which passes through
a given point and touches a given straight line is a parabola.
Ex. 15. The locus of the centre of a circle which touches a
given circle and a given straight line is a parabola.
The focus is the centre of the given circle, and the directrix a
right line parallel to the given one at a distance from it equal to
the radius of the given circle.
Ex. 16. P&p is a straight line through the focus S, cutting the
parabola in P and p. PN, pn are drawn at right angles to the
axis. Prove that AIf.An = AS^.
Ex. 17. Given the directrix and two points on the curve,
construct it. Show that, in general, two parabolas satisfy the
conditions.
Ex. 18. If from a point P of a circle, PC be drawn to the
centre C, and E be the middle point of the chord PQ drawn
parallel to a fixed diameter ACB ; then the locus of the inter-
section of CP and AR is & parabola.
The focus will be at C, and the directrix will be the tangent to
the circle at A.
PROPERTIES OF CHORDS.
Def. The chord (QQ') of a conic is the finite straight
line joining any two points (Q, Q') on the curve.
Def. A focal chord (PSp) is any chord drawn through
the focus (S).
Def. The latus rectum (LL') of a conic is the focal
chord drawn at right angles to the axis.
PAEABOLA. 7
Def. The focal distance (SP) of a point (P) on a conic
is its distance from the focus.
Def. The ordinate (PJS) of a point (P) on a conic is
the perpendicular from the point on the axis.
Def. The abscis-sa (AX) of a point (P) on a parabola,
with respect to the axis, is the portion of the axis be-
tween the vertex and the ordinate of the point
Proposition II.
The lotus rectum, of a 'parabola is equal to four tiTnes
the distaTice of the focus from the vertex (LL' = -iAS).
Let LSL' be the latus rectum. Draw LM perpendicular
to the directrix.
Since the parabola is symmetrical, with respect to the
axis, LS=L'S. Therefore
Lr=2LS=2L.V= 2XS= 4AS.
Ex. 1. Find a double ordinate of a parabola which shall be
double the latus rectum.
8 GEOMETRY OF CONICS.
Ex. 2. The radius of the circle described about the triangle
LAL'—^ latiis rectum.
Ex. 3. Find the point 0 in a given ordinate PJV, such that OR
being drawn parallel to the axis to meet the curve in R, 0N+ OR
may be the greatest possible. [OiV'=2.4*S^.]
* Proposition III.
Any focal chord of a 'parabola is divided harmoni-
cally by the curve, the focus, and the directrix.
Def. A straight line .45 is said to be divided har-
monically in 0 and 0',if it is divided internally in 0 and
externally in 0', in the same ratio, that is, if
AO:OB = AO':0'B.
Produce the focal chord PSp to meet the directrix in D,
and draw PM,pni from P, p, perpendicular to the directrix.
Then, from the similar triangles DMP, Dmp,
PD:pD = PM:pm.
But PM= PS, and pm = pS.
Therefore PD:pD = PS: pS.
Hence Pp is divided harmonically in S and D.
PAEABOLA.
Ex.1. Prove th.tj^+Jg=^.
Ex.2. I^vethat^+^=^.
Ex. 3. The semi-latus rectmn is a harmonic means between
the two segments of any focal chord of a parabola.
Ex. 4. Focal chords of a parabola are to one another as the
rectangles contained bj their segments.
Proposition IV.
TTie square of the <yrdinate of any point on a 'para-
bola is equal to the rectangle contained by the lotus
rectum, and the abscissa (PiV^ = 4J./S>. AIT).
Draw PJ/ perpendicular to the directrix, and join SP.
Then, because XS is bisected in A and produced to N,
NT" = SN^+^AS . AN. [Euc. II. 8.
But NX = PM=8P.
Therefore NX'' = SP^ = SN^+PN^. [Euc. 1. 47-
Therefore PN^- = -iAS . AX.
10 GEOMETEY OF CONICS.
Ex. 1. If PL be drawn at right angles to AP, meeting the axis
in L, NL is always equal to the latiis rectum.
Ex. 2. If a circle be described about the triangle SPN, the tan-
gent to it from A =\PK.
Ex. 3. A straight line parallel to the axis bisects PN, and meets
the curve in Q ; NQ meets a line through A at right angles to the
axis, in T. Prove that 3^1 T== 2 . PN.
Ex. 4. If SQ be parallel to AP, and QM be the ordinate of Q,
prove that SM'^=AM . AN.
Ex. 5. If 0 be any point on a double ordinate PNP", and OQ
parallel to the axis meets the curve in Q, show that
(i.) 0P.0P=4AS.0Q;
(ii.) PN:ON=OR:QR.
Ex. 6. PNF is a double ordinate of a parabola. Through Q,
another point on the curve, straight lines are drawn, one passing
through the vertex, the other parallel to the axis, cutting PP in I,
v. Prove that PN^ = Nl . M.
Ex. 7. A circle has its centre at A, and its diameter is equal to
ZAS. Show that the common chord of the circle and the parabola
bisects AS.
Ex. 8. AP, BQ are two lines at right angles to AB ; A is joined
to any point Q on BQ ; a point 0 is taken on AQ such that the
perpendicular ON on AP=BQ. Prove that the locus of 0 is a
pai'abola. [Axis, AP ; Latus rectum, AB.]
Ex. 9. PM, QN are the ordinates of the extremities of two
chords AP, AQ which are at right angles to each other. Prove
that AM. 4iV= (Latus rectum)-.
Ex. 10. The latus rectum is a mean proportional between the
double ordinates of the extremities of a focal chord. (See Prop. I.,
Ex. 16).
Ex. 11. PSp is a focal chord ; prove that AP, Ap meet the latus
rectum in points whose focal distances are equal to the ordinates of
p and P respectively. (Apply Prop. I., Ex. 16.)
Proposition Y.
The locus of the iniddle points of any system of parallel
chords of a parabola is a straight line parallel to the
aacis.
Let QQ' be one of a system of parallel chords. Draw
QM, QM perpendicular to the directrix. Draw SY
PAEABOLA. 11
perpendicular to QQ', produce YS to meet the directrix
in K, axid draw KV parallel to the axis. Then KV shall
bisect QQ'. 3o\ji KQ, KQ', SQ,B.ndi SQi.
Then MK' = KQ- - MQ' [Eua I. 47.
But KQ^^KT^+Qir^ [Euc.1.47.
and Q^ = SY'- + QT^. [Eua L 47.
Therefore ME^ = KY^-ST^.
Similarly MK^- = KQ'^ - ^rQ''-
= KY^-SY\
Therefore MK=^rK,
but, since KV is parallel to MQ and WQ', QQ is bisected
at F.
Now QQ' being fixed in direction and KSY being per-
pendicular to it, KSYvi a fixed straight line and ^ is a
fixed point. Therefore KV, which is parallel to the axis.
12 GEOMETRY OF CONICS.
is a fixed straight line bisecting all chords parallel
to QQ'.
Def. A diameter of any curve is the locus of the
middle points of a system of parallel chords drawn in
the curve.
It has just been proved that the diameters of a parabola are
straight lines. It will be shown hereafter that the diameters of the
other conies are also straight lines. It should be observed, however,
that a diameter is not necessarily a straight line for all curves.
Def. The half chords (QV, QfV) intercepted between
the diameter and the curve, are called the ordinates to
the diameter.
Def. The abscissa of a point on a parabola with re-
spect to any diameter is the portion of the diameter
intercepted between the ordinate of the point and the
parabola.
Def. In the parabola, the vertex of a diameter is the
point in which it cuts the curve.
Ex. 1. The perpendicular from the focus upon a system of
parallel chords intersects the diameter bisecting the chords upon
the directrix.
Ex. 2. If a system of parallel chords make an angle of 45° with
the axis, their diameter passes through an extremity of the latus
rectum (see Prop. IV.).
Ex. 3. A parabola being traced on paper, find its focus and
directrix.
The direction of the axis is given by the straight line joining
the middle points of a pair of parallel chords. The position of
the axis is found by observing that the middle point of any chord
at right angles to its direction lies on it. At any point JV'^on the
axis, draw a perpendicular to it NK='2AN. Join KA, cutting
the curve in L, which will be an extremity of the latus rectum.
Ex. 4. The difference between tlie segments of any focal chord
is equal to the parallel chord through the vertex.
Ex. 5. QSQf is a focal chord ; QM, Q'M' are perpendicular to
the axis. Show that MM' is equal to the parallel chord thiough
the vertex.
PARABOLA. 13
Ex. 6. AF is any chord through the vertex, and F£ is dra-wTi
at right angles to AP, meeting the axis in £J. AE is equal to the
focal chord parallel to A P.
Ex. 7. The middle points of any two chords of a parabola
equally inclined to the axis, are equidistant from the axis,
Ex. 8. If a parabola drawn through the middle points of the
sides of a triangle ABC meets the sides again in a, j8, y, the lines
Aa, Bfi, Cy will be parallel to each other. [Each is parallel to
the axis.]
Propositiox YI.
The 'parameter of any cUavieter of a parabola, is four
times the line joining the focus with the vertex of the
diaTneter.
Def. The parameter of a diameter is the length of the
focal chord bisected by the diameter.
Draw 8K at right angles to the focal chord PSp, to
meet the directrix \n K ; draw PM, pm at right angles
to the directrix, and KBV parallel to them. Then KBV
is the diameter bisecting the chord PSp (Prop. V.).
Join SB.
14 GEOMETEY OF CONICS.
Then, since K8V is a right angle, and KB = BS, we
have
KB = BS=BV,
or KV==2BS.
Now, because Pp is bisected in V,
Pp = PS+ S2:> = PM+pm
= 2KV=4BS.
Ex. 1. Given the length of a focal chord, find its position.
Ex. 2. Draw a focal chord PSp, svich that SP=3Sp.
Proposition VII.
The ordinate to any diameter of a parabola at any
point is a mean proportional to its parameter and
the abscissa of the point with respect to the diameter
{QV^ = 4>BS.BV).
Let QQ' be any chord. Draw SY at right angles to it,
and produce YS to meet the directrix in if. Draw KBV
PARABOLA. 15
parallel to the axis, so that BV is the diameter bisecting
QQ' in V, QF being the ordinate and BV the abscissa.
[Prop. V.
Draw SV parallel to QQ\ and QM, QD, V'G at right
angles to the directrix, ^Fand QQ' respectively.
Then QD^ = 2MK^
= KY-2-ST'; [Prop.V.
and, from the similar triangles QVD, KVY, and V'VC,
QD:QV=KY:KV
= V'C:V'V
=SY: V'V.
Therefore QV^ = KV^- V'VK
But as jK"F' is bisected in B, [Prop. VI.
KV^= V'V+ABV.BV. [Euc. II. 8.
Therefore QV'-=4BV.BV'
= 4BS.BV. [Prop. VI.
Ex. 1. If any chord BR meets QM and QQ" in L and y, prove
tha.tBL'^=BIf.BR
Ex. 2. If QQ' meets anv chord BR in xV, and the diameter
through R in .V, prove that Qr^= VX. VN'.
Ex. 3. If QOQ' be any chord meeting the diameter jBFin 0,
and QV, Q'V ordinates to the diameter, then BO' = BV.BV'.
Let QB produced meet the diameter through Q' in £J, and draw
ER parallel to the ordinate meeting B V produced in R.
Then QV^ :^V^ = BV'^ iBV.B V.
But QV2:BV^ = Q'V^:BR:-,
BV.BV' = BRr-',
BV:BR=BR:BV';
or BV:RV=BR:RV'.
But BV:RV=QB:QE
=BO:RV';
BO=BR.
Ex. 4. If POP be the chord bisected by the diameter BOV at
o.po^^QV.qr.
Ex. 5. Througli a given point, to draw a chord of a parabola
■which will be divided in a given ratio at the point.
Tlirough the given point 0, draw the diameter BO. Tlien if I'',
V be the feet of the ordinates drawn through the extremities of
the chord sought, it is clear that ^F' :.BF is as the square of the
16 GEOMETRY OF CONICS.
given ratio. Also, BV.BV'=BO% whence the points V, V are
known.
Ex. 6. If any diameter intersect two parallel chords, the rect-
angles under the segments of these chords are proportional to the
segments of the diameter intercepted between the chords and the
curve.
If QQ' be one of the chords meeting the diameter ^F in F, and
if 0 be its middle point,
QV.Q'V=Q(P-OV^=^S,BV.
Ex. 7 QQ is a fixed straight line, and from any point F in it,
VB is drawn in a fixed direction such that BV is proportional to
QV .QV. Show that the locus of 5 is a parabola passing through
Q, Q and having its axis parallel to B F.
Ex. 8. Given the base and area of a triangle, the locus of its
orthocentre is a parabola.
Ex. 9. BOy B'Cf are any two diameters. A line is drawn
parallel to the ordinate to BO, cutting the curve in Z), and BO,
BE, B'O in 0, C, E respectively. Prove that OD'^=OC.OE.
(Through F draw a parallel to EO.)
♦Proposition VIII.
If two chords of a parabola intersect each other, the
rectangles contained by their segments are in the ratio
of the parallel focal chords.
Let the chords QQ' and qq' intersect in a point 0
PARABOLA. 17
within the parabola. Bisect QQ' in 1', an J draw the
diameters OR, VB. Draw RW parallel to QQ'.
Then, because QQ' is bisected in T",
QO .Q'0 = Qr^-0 V- [EiK-. II. .5.
= QV2-RW'^
= 4BV. BS- 4BW. BS [Prop. VII.
^4^BS.WV
= ^BS.OR.
Similarly, if bv be the diameter bisecting qq',
q0.q'0 = ibS.OR.
Therefore QO . Q'O : qO . q'O = -^BS : ihS ;
that is, as the focal chords parallel to QQ' and qq' respec-
tively. [Prop. YI.
The proposition may be similarly proved when the
chords intersect outside the curve.
Ex. 1. If two intersecting chords be parallel to two others, the
rectangles contained by the segments of the one pair are propor-
tional to the rectangles contained by the segments of the other paii".
Ex. 2. Deduce Prop. III.
Ex. 3. Given three point.s on a parabola and the direction of
the axis, construct the curve.
Ex 4. Inscribe in a given jmrabola a triangle having its sides
pjvrallel to three given straight lines.
* Proposition IX.
If a circle intersect a parahoUi in four points their
common chords imll he equally inclined, tivo and tivo, to
the axis.
Let Q, Q', q, q be the four points of intersection.
Then QO . Q'O = qO . q'O. [Euc. III. 35.
Therefore, the focal chords parallel to QQ' and qq' are
equal to each other. [Prop. VIII.
18 GEOMETRY OF CONICS.
And they are therefore ecj^ually inclined to the axis,
from the symmetry of the figure. (See also Prop. I.,
Ex. 10.)
Therefore the chords QQ', qq are equally inclined to
the axis.
In like manner, it may be shown that the chords Qq
and q'Q\ as well as the chords Qq' and qQ', are equally-
inclined to the axis.
Ex. 1. If a circle cut a parabola in four points, two on one side
of the axis and two on the other, the sum of the ordinates of the
first two is equal to the sum of the ordinates of the other two
points. (See Prop. V., Ex. 7.)
Ex. 2. If three of the points are on the same side of the axis,
the sum of their ordinates is equal to the ordinate of the fourtlj
l)oint.
Proposition X.
If any chord QQ' of a parabola intersects the directrix
in J), SD bisects the extensor angle between SQ and SQ'.
Draw QM, Q'M' perpendicular to the directrix.
PARABOLA. 19
Then, by similar triangles,
QD:Q'D = QM.Q'^r
= 8Q:SQ'.
Therefore SD bisects the exterior angle Q'Sq. [Euc VI. A.
Ex. 1. Given the focus aud two points on a parabola, find the
directrix.
The point D, being the intersection of the chord QQ" and the
bisector of the angle Q'Sq, is on the directrix, which touches the
ciix'le descriljed with Q as centre and radius ^.S".
Ex. 2. PQ, pq are ftxal chords. Show that Pp, Qq, as also Pq,
pQ, meet on the directrix.
If they meet the directrix in A', A'', KSK' is a right angle.
Ex. 3. Given the focus and the directrix, trace tlie jmrabola by
means of this proposition. (For other constructions, see Prop. 1.,
and Ex. 2, Ex. 3.)
Determine the vertex A as the middle point of SX. Take any
ix)int D on the directrix ; make the angle BSp equal to the angle
DSA, and let pS and DA produced meet in P. P is a point on the
I)ai*abola.
Ex. 4. ^ is a point on the pai-alwla. If QA produced meet the
directrix in Z>, JfSD is a right angle.
Ex. 5. PQ is a double ordinate, and PA' cuts the curve in P :
show that the focus lies on I^Q.
Ex. 6. If two fixed points Q, Q on a parabola be joined with a
thirtl variable point 0 on the curve, the segment qi^ intercepted on
the directrix b}- the chords QO, f/0 protluced, subtends a constant
angle at the focus.
20 GEOMETRY OF CONICS.
The angle qSq' may be proved to be equal to half of the angle QS(^.
Ex. 7. If QQ' be a focal chord, the angle qSq' is a right angle,
and qX . g'A''=(semi-latus rectuni)^.
Ex. 8. Show that a straight line which meets a parabola will,
in general, meet it in two points, except when the line is parallel to
the axis, in which case it meets the curve in one point only ; and
no straight line can meet the curve in more points than two.
Let DQ' be any straight line which meets the directrix in D and
the curve in Q'. Make the angle DSq equal to the angle DSQ', and
let qS\ DQ' intersect in Q. Then since
SQ : SQ' = QD : Q'D = Q2f : Q'Hr,
§ is a point on the curve. If, however, DQ' be parallel to the axis,
qS will coincide with the axis, and D'Q' will meet the parabola in
the point Q' only (the other point of intersection in this case being
really at infinity). Again SQ, SQ', being equally inclined to DS, if
there be a third point of intersection Q", SQ, SQ" will make the
same angle with DS, which is impossible.
PROPERTIES OF TANGENTS.
Def. A tangent to a conic is the limiting position of a
chord whose two points of intersection with the curve
have become coincident.
Thus, if P and P' be two points on a conic, and if the
chord PP' be so turned about P that P' may approach P,
then in the limiting position when P' moves up to P and
coincides with it, the chord becomes the tangent to the
conic at P.
Again, if a chord PP' moves parallel to itself until P
and P' coincide at a point B on the conic, PP' becomes
in its limiting position the tangent to the curve at tlie
point B.
I
PARABOLA. 21
Hence, a tangent raaj' be said to be a straight line
which passes through two consecutive or coincident
points on the curve.
It will be seen that, generally, to a chord-property of a
conic, there coiTesponds a tangent-propert}'.
Thus, in Prop. V., if the chord QQ' moves paitillel to
itself until Q' coincides with Q at the point B on the
curve, the chord in this its limiting position becomes the
tangent to the parabola at B, which is thus seen to be
parallel to the sj'stem of chords bisected by the diameter
BV. (See Prop. XI.)
* Again, in Prop. YIII., let the chords QQ', qq intersect
at a point 0 outside the parabola. Let the chord OQQ'
be made to turn about the point 0, until Q' coincides with
Q at a point R on the curve, so that OR becomes the
tangent to the curve at the point R, and OQ, OQ' become
each equal to OR. In like manner, let Oqq be made to
turn about the point 0, until q coincides with g at a point
r on the curve, so that Or becomes the tangent to the
curve at the point r, and Oq, Oq, become each equal to
Or. Hence, we have the following proposition : —
The squares of any two intersecting tangents to a
pai-abola are in the i-atio of the parallel focal chords.
Ex. 1. If OrO' be the tangent to a parabola at T, and if OPQ^
OP(jl be a pair of parallel chords,
OT- ■.aT-=op.oq: o'P . aq.
Ex. 2. If TOO be the tangent to a parabola at T, CfP a tangent
from 0", and OPQ a chord jjarallel to (yP^ cutting the chord of con-
tact Pq in R, prove that OP. OQ=OBr.
From Ex. 1,
OP.OQ: OT-=aP^ : OfT-==OB?\OT^.
Cf. Prop. XXI., Ex. 8.
* Next, in Prop. IX., suppose q to coincide with Q, and
22 GEOMETRY OF CONICS.
therefore also with 0 ; then the circle and the parabola
will touch each other at 0, the chords OQ', oq being
equally inclined to the axis. Hence
If two chords OP, OQ of a parabola are equally inclined
to the axis, the circle round OPQ touches the parabola
at 0.
Ex. If one of the cliovds OP Ije at right angles to the tangent
to the curve at 0, the angle OQP is a right angle.
Similarly, if a circle touches a parabola at 0 and cuts
it again in P and Q, the tangent at 0 and PQ are equally
inclined to the axis.
Ex. If a circle touches a parabola at 0 and cuts it in P and Q,
and PU, ^F parallel to the axis meet the circle in U, V, show that
r^'l'is parallel to the tangent at 0.
Again, consider Prop. X. Let the chord QQ' be made
to turn about Q, until Q' coincides with Q, so that the
chord becomes the tangent to the parabola at the point
Q. The angle QSQ' vanishes, and, therefore, the exterior
angle Q';S'(/ becomes equal to two right angles. But
since 8D always bisects the angle Q'Sq, SD will, in this
limiting position, be at right angles to SQ. Hence the
following proposition : —
The tangent to a parabola from any point on the
directrix, subtends a right angle at the focus. (See
Prop. XII.)
Def. A circle or a conic is said to touch a conic at a
point P when they have a common tangent at that
point.
PROPOSITION XI.
'The tangent to a ijarahola at its 2>oint of intersection
PARABOLA.
uith a diameter is parallel to the system of chonU
bisected hy the diameter
Let BY be the diameter bisecting a system of chords
]iarallel to QQ'.
Let Q(^ be made to move parallel to itself, so that Q
may coincide with V. Since QV is always equal ix) Q'V
(Prop, v.), it is clear that Q' will also coincide with B, or,
the chord in this, its limiting position, will be the
tangent to the pai~abola at B.
Ex. Draw a tangent to a parabola mikiug a given angle with
the axis.
Proposition XIL
Th£ poii;ion of the tantjent to a imrabola at any jyoint,
intercepted between that point and the directrix, subtend"^
o right angle at the focus.
Let any chord QQ' of the parabola intersect the directrix
in Z.
Then SZ bisects the exterior angle Q'Sq. [Prop. X.
24
GEOMETRY OF CONICS.
Now, let the chord Q(^ be made to turn about Q until
the point Q' moves up to and coincides with Q, so that
the chord becomes the tangent to the parabola at Q. lu
this limiting position of the chord QQ\ since Q and Q'
coincide, the angle QSQ' vanishes, and therefore the
angle Q'tSq becomes equal to two right angles. But since
PARABOLA.
2o
SZ always bisects the angle Q'Sq, in this case the angle
QSZ is a right angle.
Ex. 1. If a line QZ meeting the curve in Q and the directrix in
Z, subtend a right angle at the focus, it will be the tangent to the
curve at Q.
Ex. 2. The tangents at the extremities of the latus rectum meet
the directrix on the axis produced.
* Proposition XIIL
If from any point 0 on the tangent at P of a parabola
perpendiculars OU and 01 he drawn to SP and the
directrix respectively, then
SU=Ol
Join /S'Z, and draw PJI perpendicular to the directrix.
Because ZSP is a right angle, [Prop. XII.
ZS is parallel to OU.
Therefore, by similar triangles,
SU:SP = ZO:ZP
= OI:PM.
But SP = PJiT;
therefore SU= 01.
26 GEOMETRY OF CONICS.
This ])ro|)eity of the parabola is the partioulai- case of a <feiieral
property of all coiiics discovered by Adams.
Ex. If a line OP meet the parabola at P, and 01, OU being
drawn at right angles to the directrix and >S'P respectively, 8U= 01,
then OP will be the tangent to the curve .it P.
Proposition XIV.
Tlte tangent at any point of a parabola bisects the
angle which the focal distance of the point makes with
the perpendicular drctivn from the point on the directrix,
and conversely.
Let the tangent at the point P meet the directrix in Z.
Draw PM perpendicular to the directrix, and join 8P, *S'Z.
Then, since the angle PSZ is a right angle, [Prop. XII.
>ST-^ + SZ-' = PZ\ ^ [Buc. I. 47.
Also PJ/^ + J/Z'' = PZ-^ ; [^^"c. 1. 47.
therefore SP'^ + SZ'^ = PM- 4- MZ~.
But SP = P]\I;
therefore SZ=MZ.
Now, in the two triangles ZPM, ZPlS, the two sides
PM, MZ are respectively equal to the two sides SP, SZ,
i
PAEABOLA. ^7
and the side PZ is common ; therefore the two triangles
are equal, and the angle SPZ is equal to the angle MPZ,
that is, PZ bisects the angle SPM.
Conversely, if PZ bisects the angle SPM, PZ is the
tangent at P. For, if not, and if possible, let any other
line PZ' be the tangent at P, then by what has been
proved PZ' will bisect the angle SPM, which is im-
possible ; therefore PZ is the t<ingent at P.
Xote. — It may be shown from the definition of the parabola that
the straight line which bisects the angle between *S'P and PJI can-
not meet the cur\-e again in any other point ; hence PZ would alf>(»
be the tangent to the parabola at P, according to Euclid's definition
of a tangent.
Corollary. — The tangent at the vertex of a parabola is at right
ingles to the axis.
Ex. 1. Show how to di-aw the tangent at a given point of u
jKii-abola.
Ex. 2. Draw a tangent to a parabola making a given angle with
the axis.
Ex. 3. If the tangent at P meets the axis in T, SP=ST.
Ex. 4. Two parabolas have the same focus, and their axes in the
same straight line, but in opposite directions. Prove that they
intersect at right angles.
Xote. — Two curves are said to intersect at right angles when
their tangents at a common point are at right angles.
Ex. 5. Given the vertex of a diameter of a parabola and a
corresponding double oixlinate, construct the curve. (Applv
Prop. VII.)
Ex. 6. If ZP l)e produced to R, the angles SPR and MPIt are
equal.
Ex. 7. PZ bisects SM nt right angles,
Ex. 8. Any point 0 on the tangent at P is equidistant from
J/'and S.
Ex. 9. If the tangents to the parabola at Q and (/ meet in 0,
and QM, Q'M' be the perpendiculars on the directrix from Q and Q',
OM, OS, UM' are all equal.
Hence deduce, by analysis, the construction for Prop. XVII..
namely, to draw two tangents to a parabola from an external
point 0.
Ex. 10. The tangent at any point of a parabola meets the
directrix and the latus rectum in two ]x>ints equidistant from the
focus.
28 GEOMETRY OF CONICS.
Ex, 11. The focal distance of any point on a parabola is equal to
the length of the ordinate of tliat point produced to meet the
tangent at the end of the latus rectum. (See Prop. XII., Ex. 2.)
Ex. 12. 0 is a point on the tangent at P, such that the perpeu- "
dicular from 0 on SP is equal to ^AS ; find the locus of 0. (A
])arabola of which the vertex is on the directrix of the given one.
Apply Prop. VII., Ex. 7.)
Ex. 13. If a leaf of a book be folded so that one comer moves
along an opposite side, the line of the crease touches a parabola.
Let the leaf BCXS be so folded that S coincides with a point J/
on CX; let the crease 7T' meet XS, £S in T, T' respectively.
Draw MP at right angles to CX, meeting TT' in P ; join SP.
Then SP==PM, lSPT=-MPT; TT\ therefore, touches at P a
parabola, of which the focus is S and directrix C.
Def. The portion of the axis intercepted between the
tangent at any point of a conic and the ordinate of that
point is called the sitbtangent
* Proposition XV.
The subtangent of any point of a parahola is bisected
at the vertex, that is, is equal to double the abscissa of the
point with respect to the axis.
M
P^^
///
T X
aI S N
Let the tangent PT at P meet the axis in T. Draw
PARABOLA. 29
PX, PM perpendicular to the axis and directrix re-
spectively.
Then, the angle >STP = the angle TPM
= the angle TPS. [Prop. XIV.
Therefore ST=^ SP= PJ/= XX
But AS=AX.
Therefore AT=AX,
or XT=2^X.
Ex. 1. If T is the middle point of AX, prove that X is the
middle point of AS.
Ex. 2. The radins of the circle descril«tl round the triangle TrX
U^f{SP.AX).
Ex 3. The locus of the midfUe ix)ints of the focal chords of a
jjarabola is another jmrabola having the same axis and passing
through the focus. (Apply Prop. VII., Ex. 7.)
Ex. 4. The diameter through P meets at E, a right line through
S parallel to the tangent at J\ Prove that the locus of E is a
paralx:>la.
If En be perpendicular to the axis, nS=XT=2AX. If S' be
taken on the axis, such that 2SS' = AS, the relation rX- = 4AS. AX
gives En'- = -hSS'.S7i, showing the locus to be a paralwla whose axis
coincides with that of the original one, whose vertex is at S, and
latus rectum half that of the original jjarabola.
Ex. 5. If SM meets FT in T, XY= TY.
Ex. 6. If the tangent at P meets the tangent at the vertex in
T,AT^=AS.AX.
Ex. 7. If SE be the perpendicular from >S^on the line tlirough P
at right angles to PT, show that SE^- = AX.SP. {•2SE=PT.
Apply Prop. IV.)
Ex. 8. Given the vertex, a tangent ami its point of contact, con-
struct the cui"ve.
Produce PA to P, such that AP = AP ; if the circle on J/^ as
diameter meets the tangent at /' in T, TA is the axis. Then applv
Prop. XIV.
Ex. 9. Find the locus of the intei-section of the perpendicular
from the vertex on the tangent at any point with the diameter
through that point, (A right line pai-allel to the du-ectrix. Applv
Prop. IV.)
so
GEOMETRY OF CONICS.
* Proposition XVI.
The tangents at the extremities of a focal chord of a
'parabola intersect at right angles on the directrix.
Draw ^Z at right angles to the focal chord PSp, meet
ing the directrix in Z Join PZ, pZ, and draw PM, imi
l)erpendiculars to the directrix.
Then ZP-^=^ZS^ + SP-'
= ZM~-\-PM\ [Euc. I. 47.
Bat SP = PM.
Therefore ZS = ZM
Therefore from the triangles Z^P and ZMP, the angle
/5'PZ=the angle MPZ, and the angle SZP = the angle
MZP. [Euc. T. 8.
Similarly,
the angle SpZ= the angle mpZ,
and the angle SZp — the angle mZp.
Therefore, l^Z and pZ are the tangents at P and p.
[Prop. XI V^.
PAEABOLA. 31
Also,
the angle PZp = ^ the angle MZS+ i the angle mZS
= one right angle
Ex. 1. Show tliat Min is bisected in Z.
Ex. 2. If two tungeiits be drawn to a pai-alM)]a from any point
on the directrix, they shall be at right angles.
Ex. 3. If perpendiculars through P, p, to ZP, Zp respectively,
meet in 0, the distance of 0 from the directrix varies as PS.pS.
Apply Prop. III., Ex. 4.)
Ex. 4. Find the locus of 0 in Ex. 3. [A i>ai-abola having the
same axis as the given one.]
Ex. 5. Show that the circle de.scribetl on the focal chord Pp a.s
•liameter touclies the directrix at Z.
Ex. 6. If a circle described upon a chord of a pai-abola as
diameter meets the directrix, it also touches it ; and all chords for
which this is possible, intersect in a fixed ixjint. [Tlie f<x3us.]
The distance of the middle point of the chord from the directrix
is always greater than half the chord, unless the chord passes
through the focus.
Ex. 7. Tangents at the extremities of a focal chord cut off equal
intercepts on the latus rectum. (Apply Prop. XIV., Ex. 10.)
Ex. 8. Prove that SM, Sm are respectively j>arallel to Zp, ZP.
Ex. 9. The locus of the intersection of any two tangents to a
parabola at right angles to each other, is the directrix.
Ex. 10. Given two tangents at right angles, and their points of
contact, construct the cur\-e.
Proposition XVII.
To draw two tangents to a imrahoUi jroni an external
point.
Let 0 be the external point. With centre 0 and
radius OS, describe a circle cutting the directrix in M
and M'. Draw MQ, M'Q' at right angles to the directrix
to meet the parabola in Q and Q. Join OQ and OQf;
these shall be the tangents required.
Join OS, OM, 021' SQ and SQ^
32 GEOMETEY OF COXICS.
Then, in the triangles OQM, OQS, the sides MQ, QO are
equal to the sides SQ, QO respectively, and OM is equal
to OS. Therefore the angles OQM, OQS are equal.
Therefore OQ is the tangent to the parabola at Q.
[Prop. XIV.
Similarly, OQ' is the tangent at Q'.
Note. — For an analysis of the construction, see Prop. XIV., Ex. 9.
It should be observed that in order that the construction may be
possible, the circle described with 0 as centie and with radius OS
must meet the directrix, that is, the distance of 0 from S must be
either greater than or equal to its distance from the directrix.
The former is the case when the point is outside the parabola
(Prop. I., Ex. 7) ; and as in this case the circle must intersect the
directrix in two points only, it follows that two tangent.s, and no
more, can be drawn to a parabola from an external point. In the
second case the point 0 is evidently on the pai-abola, and the circle
touches tlie directrix, that is, meets it in two coincident points ;
the two tangents in this case coincide, that is, only one tangent can
be drawn to a parabola at a given point on it. The distance of
any point inside the parabola being less than its distance from
the directrix (Prop. I., Ex. 6), no tangent can be drawn to a
parabola from any point within it.
PAKABOLA.
33
Ex. 1. If the point 0 be on the directrix, show from the
construction that the tangents intersect at right angles.
Ex. 2. If 0 be on the axis produced, at a distance from the
vertex A = ^AS, the figure OQSQ will be a rhombus.
Ex. 3. Alternative Construction. — With the given point 0 as
centre and radius OS, describe a circle cutting the directrix in
M and J/'. The perpendiculai-s from 0 upon SM and SM' will,
when produced, touch the curve. (See Prop. I., Ex. 3.)
Ex. 4 Alternative Constructimi. — In the figure of Prop. XIII.,
taking 0 as the given point, draw 01 at right angles to the
directrix. With centre S and radius equal to 01, describe a circle ;
and from 0 draw OU and OU' tangents to this circle. SU, SU'
produced will meet the parabola in the poiuts of contact of the
tangents from 0. (See Prop. XIII., Ex.)
For another alternative construction, see Prop. XXIIL, Ex. 13.
Proposition XVIII.
The two tangents OQ, OQf of a parabola subtend eqital
angles at the focus ; and the triangles SOQ, SOQ' are
similar.
With centre 0 aad radius OS, describe a circle cutting
c
34 GEOMETEY OF CONICS.
the directrix in 31 and M'; draw MQ, M'Q' at right angles
to the directrix to meet the curve in Q, Q'. Then OQ and
OQ' are the tangents to the curve from 0. [Prop. XVII.
Join OM, OM, OS, SQ, SQ', and S3f, cutting OQ in F.
In the two triangles il/QFand SQY, the sides MQ, QY
are equal to the sides SQ, QY, and the angles MQY, SQY
are equal ; [Prop. XIV.
therefore the two triangles are equal in every respect ; and
the angles MYQ, 8YQ are equal, each being thus equal to
a right angle. [Euc. I. 4.
Now, the angle >SfQO = the angle 3/QO,
and the angle MQO = the angle SMM',
each being the complement of the angle QMY.
Therefore
the angle SQO = the angle SMM.
But the angle SMM' = ^ the angle SOM', [Euc. III. 20.
and from the equality of the triangles SOQ', M'OQ',
[Prop. XVII.
the angle ;SfOQ'==the angle M'OQ',
or, the angle SOQ'= ^ the angle SOM'.
Therefore the angle SQO = the angle SOQf.
Similarly, the angles QOS and OQ'S are equal, as also
the remaining angles QSO, Q'SO.
Therefore the two triangles SOQ, SOQ' are similar.
Ex. 1. Prove that
(i.) SQ.SQ'=^SO^; (ii.) OQ^: OQ^^^^SQ -.SQ'.
Ex. 2. If two tangents drawn from any point on the axis be cut
by any third tangent, the points of intersection are equidistant
from the focus.
Ex. 3. The angle subtended at the focus by the segment inteiv
cepted on a variable tangent by two fixed tangents, is constant.
Ex. 4. OS and a line through 0 parallel to the axis make equal
angles with the tangents.
PARABOLA. 35
Ex. 5. The straight line bisecting the angle QO^ meets the axis
in R ; prove that SU—SR.
Ex. 6. If two tangents drawn from any point on the axis be cut
bv a third tangent, their alternate segments are equal. (Cf. Prop.
XXI., Ex. 10.)
Ex. 7. If the tangent and normal at any point P of a parabola
meet the tangent at the vertex in K and L respectively, prove that
KU:Sr-=SP-AS:AS.
Ex. 8. If from any point on a given tangent to a parabola, tan-
gents be drawn to the curve, the angles which these tangents make
with the focal distances of the points from which they are drawn,
are all equal.
Each angle is equal to the angle between the given tangent and
the focal distance of the point of contact.
Ex. 9. Of the two tangents drawn to a parabola from any point,
one makes with the axis the same angle as the other makes with
the focal distance of the point.
Ex. 10. Two parabolas have the same focus and axis, with their
vertices on the same side of their common focus. Tangents are
drawn from any point P on the outer parabola to the inner one.
Show that they are equally inclined to the tangent at P to the
outer curve. (Apply Ex. 9, and Prop. XIY.)
Ex. 11. If the tangent at any point R meets OQj 0^ in y, j*,
show that Qq:qO=0^ : yV = qR : Rq'.
[The triangles OqS, Rq'S are similar.]
Ex. 12. If tangents be drawn from any point on the latus rec-
tum, show that the semi-latus-rectum is a geometric mean between
the ordinates of the points of contact. (Apply Prop. I., Ex. 16, and
Prop. IV.)
Ex. 13. If Pr, PT be two diameters, and PY, PV ordinates
to these diameters, show that PT'=Pr'. (Applv Prop. VII. and
Ex. 1.)
Ex. 14. If one side of a triangle be parallel to the axis of a par-
abola, the other sides will be in the ratio of the tangents parallel
to them.
Proposition XIX
The eoderior angle between any two tangents to a
parabola is equal to the angle ivhich either of them
subtends at the focus.
3G GEOMETRY OF CONICS.
Let OQ and OQ' be the two tangents, and S the focus.
Join SO, SQ, and SQ'.
The angle SOQ' = the angle SQO. [Prop. XYIII.
To each of these equals add the angle SOQ ; therefore the
angles SOQ and SQO are together equal to the angle
QOQ'. But the exterior angle HOQ' is the supplement of
the angle QOQ' (Euc. I. 13), and the angle OSQ is the
supplement of the angles SOQ and SQO (Euc. I. 32),
Therefore
the angle HOQ' = the angle OSQ
= the angle OSQ'. [Prop. XVIII.
Ex. 1. Two tangents to a parabola, and the points of contact of
one of them being given, prove that the locus of the focus is a circle.
The circle may be shown to pass through the given point of con-
tact and the intersection of the tangents, and to touch one of them.
Ex. 2. If a parabola touch the sides of an eqi;ilateral triangle,
the focal distance of any vertex of the triangle passes through the
point of contact of the opposite side.
Ex. 3. Given the base AB and the A-ertical angle Cof a triangle
ACB, j5nd the locus of the focus of a parabola touching CA, CB in
A and B.
Ex. 4. EiB the centre of the circle described about the triangle
PARABOLA. 37
OQ^ I prove that the circle described about the triangle QECjf
passes through the focus.
Ex. 5. A circle passing through the focus cuts the parabola in
two points. Prove that the exterior angle between the tangents to
the circle at those points is four times the complement of the exterior
angle between the tangents to the parabola at the same points.
* Proposition XX.
The circle cvrcumscribing the triangle foi'med hy any
three tangents to a 'parabola passes through the focus.
Let the three tangents at the points P, Q, R form the
triangle pqr.
Join SP, Sp, Sq, Sr.
The angle Srp = the angle SPr, [Prop. XVIII.
and the angle Sqp = the angle SPr ; [Prop. XYIII.
therefore the angle Srp = the angle Sqp.
Therefore the points p, q, r, S lie on a circle, or the
circle round the triangle pqr passes through the focus.
Ex. 1. What is the locus of the focus of a parabola which
touches three given straight lines ?
38 GEOMETEY OF CONICS.
Ex. 2. A parabola touches each of four straight lines given in
position. Determine its focus.
The four circles circumscribing the iowr triangles formed by the
given straight lines, will intersect in the same point, namely, the
focus required. Hence, the curve may be described. (See Prop.
XXIIL, Ex. 5.)
Ex. 3. If through p, q, r lines be drawn at right angles to *S^,
JSq, Sr respectively, they will meet in a point.
Ex. 4. Prove that the orthocentre of the triangle pqr lies on
the directrix. (Apply Prop. XII.)
* Proposition XXI.
If through the point of intersection of two tangents to
a parabola a straight line be drawn parallel to the axis,
it will bisect the chord of contact.
Let OQ and OQ' be the two tangents, and let OV
drawn parallel to the axis meet QQ' in V and the
directrix in R. Draw Q3I and Q'M' perpendicular to
the directrix, and join OS, OM, OM'.
PAEABOLA. 39
Then 0M= 0S= OM', [Prop. XVII.
and OR, which is drawn at right angles to the base of
the isosceles triangle OMM', bisects it.
Therefore MR^M'R.
But since MQ, RV, M'Q are parallel to one another,
QV:QV=MR:M'R-,
therefore QV=Q'V,
or, QQ' is bisected in V.
Ex. 1. The tangents at the extremities of any chord of a
parabola meet on the diameter bisecting that chord.
Ex. 2. The circle on any focal chord as diameter touches the
directrix.
Ex. 3. The straight lines drawn through the extremities of a
focal chord at right angles to the tangents at those points, meet on
the diameter bisecting the chord-
Ex. 4. Given two tangents and their points of contact, find the
focus and directrix.
Ex. 5. Given two points P, ^ on a parabola, the tangent at one
of the points P, and the direction of the axis, construct the curve.
If the tangent at P meets the diameter bisecting PQ in T, TQ is
the tangent at Q. Hence the focus by Prop. XIV.
Ex. 6. If a line be drawn parallel to the chord of contact of two
tangents, the parts intercepted on it between the curve and the
tangents are equal.
Ex. 7. OP, OQ are two tangents to a parabola, and V is the
middle point of PQ. Prove that OP. 0Q=20S.0V.
On QO produced take Oq = OQ; then apply Prop. XYIII. to
show that the triangles POQ and OSQ are similar.
Ex. 8. If from any point 0 a tangent OT'and a chord OPQ be
drawn, and if the diameter TR meet the chord in R, prove that
OP. OQ = OR-. (Cf. Tangent Properties, Ex. 1, 2.)
Draw the tangent KO'P' parallel to the chord, meeting RT va. K,
OT produced in 0, and the curves in P'. Draw the diameter &II
bisecting TP', so that 0'P' = K&. Then
OP. OQ : OT-=-aP"^ : aT^ = OK^ : aT^^OR* : OT^.
Ex. 9. Given a chord PQ of a parabola in magnitude and
position, and the point R in which the axis cuts the chord, the
locus of the vertex is a circle.
If the tangent at the vertex meets PQ in 0, OP.OQ = OB?.
.'. 0 is a fixed point ; OR=PR. RQI{PR - RQ).
40 GEOMETRY OF CONICS.
Ex. 10. The tangents from an external point are divided by
any third into segments having the same ratio.
In fig. Prop. XX., draw the diameters r/, QQ, qq'^ pp\ meeting
PR in r', Q', q', p'. Then
Pr : rq=rQ : Qp=qp : pR.
(Cf. Prop. XVIII., Ex. 11.)
Ex. 11. The tangent parallel to QQ' bisects OQ, OQ'.
Ex. 12. 1( E be the centre of the circle through 0, Q, Q", OB
subtends a right angle at S. (Apply Prop. XX., and Ex. 11.)
Ex. 13. If OQQ' be a right angle and QJV the ordinate of Q,
prove that QQ':OQ=QN:A K
(Cf. Prop. XVI.)
* Proposition XXII,
If QV is the ordinate of a diameter PV of a parabola,
and the tangent at Q meets VP produced in 0, then OP
shall he equal to PV.
Let the tangent at P meet OQ in i^; through li draw
the diameter RW, meeting PQ in W.
Then, since MP, RQ are a pair of tangents,
QW=PW. [Prop. XXI.
Also, RP is parallel to QF; [Prop. XL
PARAJBOLA.
therefore
OP:PV=^OR:RQ
= PW:WQ.
But
PW=^WQ;
therefore
OP = PV.
41
Ex. 1. Tangents at the extremities of all parallel chords meet
on the same straight line. (Cf. Prop. XXI., Ex. 1.)
Ex. 2. Given a tangent and a point on the curve, find the locus
of the foot of the ordinate of the point of contact of the tangent,
with respect to the diameter through the given point. [A right
line parallel to the tangent.]
Ex. 3. If 0 V= Q V, 0 is on the directrix.
Ex. 4. If the diameter PV meets the directrix in 0, and the
chord drawn through the focus parallel to the tangent at P in V,
prove that VP= OP.
Ex. 5. If OQ, OQ' be a pair of tangents to a parabola, and OQQ
be a right angle, OQ will be bisected by the directrix.
Draw the diameter OP V and the tangent at P. (See Prop. XVI.,
Ex. 9.)
Ex. 6. If §F be an ordinate to the diameter PV, and pv meet-
ing PQ in V be the diameter bisecting PQ, prove that PV=ipv.
Ex. 7. PQ, PR are any two chords ; they meet the diameters
through E and Q in F and K Show that EF is parallel to the
tangent at P.
Ex. 8. If from the point of contact of a tangent a chord be
drawn, and any line parallel to the axis be drawn meeting the
tangent, curve, and chord, this line wUl be divided by them in the
same ratio as it divides the chord.
Let the diameter EB V bisecting the chord Q(^ in V meet the
tangent at Q in R. Draw the line rbv parallel to the axis, cutting
the curve and chord in h and v. Then
Qv.vr^QV: VR
= QV:2VB.
But g r2 = iBS .BV; (Prop. VII.)
QV:2BV=2SB:QV;
Qv.QV=2SB.vr.
Also Qv.Q'v = 4SB . vb ; (Prop. VIIL)
•*. _ Qv. Qv=rb :bv.
This is a generalisation of Prop. XXII.
Ex. 9. Through a given point within a parabola, draw a chord
which shall be divided in a given ratio at that point.
42 GEOMETRY OF CONICS.
Proposition XXIII.
The locus of the foot of the perpendicular from the
focus upon any tangent to a parabola is the tangent at
the vertex.
Draw SY perpendicular to the tangent at P, meeting
it in Y. It is required to show that Y lies on the
tangent to the parabola at the vertex.
Draw PJf perpendicular to the directrix, and join
MY, AY.
Now, in the two triangles MPY, SPY, the sides MP,
PF are equal to the sides SP, PY respectively, and the
angle MP F= the angle SP Y [Prop. XIV.
Therefore the angle PrJf=the angle PYS
= one right angle ; [Euc. I. 4.
therefore MY and YS are in the same straight line.
[Euc. I. 14.
Now, since SY= YM,
and SA = AX,
AY ia parallel to MX, [Euc. VI. 2.
I
PAEABOLA. 43.
and is, therefore, the tangent to the parabola at the
vertex. [Prop. XIV., Cor.
Ex. 1. Show that ST^=AS.SP. [The triangles STFy ST A
are similar. See Prop. XYIII., Ex. 1.]
Ex. 2. Show that SM is bisected at right angles by the tangent
at P.
Ex. 3. If the tangent at P meet the axis in T, and PN be the
ordinate of P, prove that PT .TY=XT. TS.
Ex. 4. If the vertex of a right angle, one leg of which always
passes through a fixed point, moves along a fixed right line, the
other leg will always touch a parabola.
The fixed point will be the focus, and the fixed right line the
tangent at the vertex, whence the directrix is known-
Ex. 5. Given two tangents and the focus of a parabola, find the
directrix.
The line joining the feet of the perpendiculars from the focus on
the given tangents, is clearly the tangent at the vertex.
Ex. 6. Prove that straight lines perpendicular to the tangents
of a parabola through the points where they meet a given fixed line
parallel to the directrix, touch a confocal parabola.
Ex 7. The focus and a tangent being given, the locus of the
vertex is a circle.
Ex. 8. Given a tangent and the vertex, find the locus of the
focus. [A parabola, of which A is the vertex and the axis the
perpendicular through A on the tangent. Apply Prop. VII., Ex. 7.]
Ex. 9. Tlie circle described on any focal distance as diameter,
touches the tangent at the vertex.
Ex. 10. PSp is a focal chord ; prove that the length of the com-
mon tangent of the circles described on Sp, SP as diameters, is
^{AS.Pp).
Ex. 11. Prove that
(i.) PT.PZ=PS^',
(ii.) PY.YZ=AS.SP.
Ex. 12. A circle is described on the latus rectum as diameter;
PQ touches the parabola at P and the circle at Q ; show that SP
SQ'a.re each inclined to the latus rectum at an angle of 30".
Ex. 13. Alternative Construction for Prop. XVII.
Let 0 be the external point ; on OS as diameter describe a circle ;
the lines joining 0 with the points of intersection of this circle with
the tangent at the vertex, will be the required tangents,
Ex. 14. In the figure of Prop. VII., prove that QII^=A:AS.BV.
Let the tangent at B meet the axis in T, and the tangent at ^ in
44 GEOMETRY OF CONICS.
Y. Then SVZ is a right angle, and the triangles QDVy VAT a,re
similar (Prop, XI.)
.-. §Z)2 ; $ 72= YA^ : YT^- = AS : TS=AS : JSaS.
But QD'^ = ABS.BV. (Prop. VII.)
QV^==AAS.BV.
Ex. 15. Given the focus and two tangents, construct the curve.
[Ex. 53.
Ex. 16. Given the focus, axis and a tangent, construct the par-
abola.
Ex. 17. Given the focus, a point P on the parabola, and the
length of the perpendicular from the focus on the tangent at P,
construct the curve.
Ex. 18. Given the focus, a tangent, and the length of the latus
rectum, construct the curve.
Ex. 19. If a parabola roll upon another equal parabola, the
vertices originally coinciding, the focus of the one traces out the
directrix of the other. [The line joining the foci in any position
cuts at right angles the common tangent.]
PROPERTIES OF NORMALS.
Def. The straight line which is drawn through any
point on a conic at right angles to the tangent at that
point is called the normal at that point.
Def. The portion of the axis intercepted between the
normal at any point of a conic and the ordinate of that
point is called the 8uhno7'mal.
Proposition XXIV.
The normal at any poiiit of a parabola makes equal
angles xvith the focal distance and the axis.
Let the normal PG and the tangent PT at any point
P on the parabola meet the axis in G and T respectively.
Join SP and draw PM perpendicular to the directrix.
Then the angle 8PT=ihe angle TPM [Prop. XIV.
= the angle STP. [Euc I. 29.
PAEABOLA. 45
But the angle TPG being a right angle is equal to the
sum of the angle STP and SGP. [Euc. L 32.
Therefore the angle SPG = ihe angle SGP.
Ex. 1. Prove that ^T=SP=SG.
Ex. 2. The normal at anv poiut bisects the interior angle be-
tween the focal distance and the diameter through that point.
Ex. 3. The focus is equidistant from FT and the straight line
through G parallel to FT.
Ex. 4. From the points where the normals to a parabola meet
the axis, lines are drawn at right angles to the normals; show
that these lines touch an equal confocal parabola.
Ex. 5. A chord FQ of a parabola is normal to the curve at P,
and subtends a right angle at S ; show that SQ = 2SF.
Ex. 6. Prove that SM and FT bisect each other at right angles.
Ex. 7. If the triangle SFG is equilateral, TG subtends a right
angle at M.
Ex. 8. Prove that the points S, F, M, Z lie on a circle which
touches FG at F.
Ex. 9. If in Ex. 8 the radius of the circle is equal to MZ, the
triangle SFG is equilateral.
Ex, 10. FSp is a focal chord ; pG is the normal at p ; GHia
perpendicular on the tangent at F. Prove that M lies on the
latus rectum. (Cf. Prop. XIY., Ex. 10.)
Ex. 11. If FF, FH be drawn to the axis, maJdng equal angles
46 GEOMETRY OF CONICS.
^ith the normal PG, prove that SCP=SF.SH. [The triangles
SPF, SHP are similar.]
Ex. 12. If 8T, SZhe perpendicular to the tangent and normal
at P respectively, prove that YZ is a diameter.
Proposition XXV".
The subnormal of any point of a parabola is equal to
half the lotus rectum.
M
i ^'^'
X
1
i
/ 1
\ ■
•A S A' G
Let the normal FG at P meet the axis in 0. Draw
PM, P]\^ perpendicular to the directrix and axis respec-
tively. Join SP.
Then, the angle SPG = the angle SGP. [Prop. XXIV.
Therefore SG=SP = PM = NX.
Therefore JSrG = XS=2AS=^ latus rectum. [Prop. 11.
The subnormal is therefore of constant length.
Ex. 1. If the triangle iSPO is equilateral, SP is equal to the
latus rectum.
Ex. 2. Show how to draw the normal at any given point with-
out drawing the tangent.
Ex. 3. If the ordinate of a point Q bisect the subnormal of a
PAEABOLA. 47
point P, the ordinate of Q is equal to the normal at P. (Apply
Prop. IV.)
Ex. 4 Prove that PCT- =AAS. SP.
Ex. 5. If C be the middle point of SG, prove that
CA'2-CP2=4i6«.
Ex. 6. If PL perpendicular to AP meets the axis in L, prove
that GL = 2AS.
Ex. 7. TP, TQ are tangents to a given circle at P and Q. Con-
struct a parabola which shall touch TP in P and have TQ for axis.
Ex. 8. The locus of the foot of the perpendicular from the focus
on the normal is a parabola.
[Apply Prop. IV. SG is the axis, the vertex is at S, the latus
rectum = .4 <S'.]
Ex. 9. If GK be drawn perpendicular to SP, prove that
PK=2AS.
Ex. 10. Pp is a chord perpendicular to the axis ; the perpen-
dicular from p on the tangent at P meets the diameter through P
in R ; prove that RP='iAS, and find the locus of P.
[The triangles PXG, Rpp are similar. The locus of ^ is an
equal parabola, having its vertex A! on the opposite side of X,
such that XI' = 4^-S'.]
Ex. 11. A circle described on a given chord of a parabola as
diameter cuts the curve again in two points ; if these points be
joined, the portion of the axis intercepted by the two chords is
equal to the latus rectum.
Show also that, if the given chord is fixed in direction, the
length of the line joining the middle points of the chords is
constant.
[Apply Prop. VIII. The middle points of the chords are equi-
distant from the axis.]
Miscellaneous Examples on the Parabola.
1. Find the locus of the point of intersection of any-
tangent to a parabola, with the line drawn from the
focus, making a constant angle with the tangent.
2. OQ, 0Q[ are tangents to a parabola ; F is the middle
point of QQ' ; OV meets the directrix in K, and QQ' meets
the axis in iV. Prove that OKNS is a parallelogram.
48 GEOMETRY OF CONICS.
3. Inscribe in a given parabola a triangle having its
sides parallel to those of a given triangle.
4. Inscribe a circle in the segment of a parabola cut off
by a double ordinate.
5. PGQ is a normal chord of a parabola, meeting the
axis in G. Prove that the distance of G from the vertex,
the ordinates of P and Q, and the latus rectum are four
proportionals.
6. If AR, SY are perpendiculars from the vertex and
focus upon any tangent, prove that
SY^ = SY.AR+SA^
7. Describe a parabola touching three given straight
lines and having its focus on another given line.
8. OP, OQ are tangents to a parabola at the points
P, Q. If SP + SQ is constant, prove that the locus of 0
is a parabola, and find its latus rectum.
9. Through any point on a parabola two chords are
drawn, equally inclined to the tangent there ; show that
their lengths are proportional to the portions of their
diameters intercepted between them and the curve.
10. The focal chord PSp is bisected at right angles by
a line which meets the axis in 0 ; show that Pp = 2 . SO.
11. On a tangent are taken two points equidistant from
the focus ; prove that the other tangents drawn from these
points will intersect on the axis.
12. The locus of the centre of the circle circumscribing
the triangle formed by two fixed tangents and any third
tangent is a right line.
] 3. A chord PQ is normal to the parabola at P, and
subtends a right angle at the vertex; prove that
BQ^S.SP.
PARABOLA. 49
14. Given the vertex, a tangent, and the latus rectum,
construct the parabola.
15. P, Q are variable points on the sides AC, AB of a
given triangle, such that AP : PC—BQ : QA. Prove that
PQ touches a parabola.
16. Apply properties of the parabola to prove that —
(i.) In any triangle the feet of the three perpendiculars
from any point of the circumscribing circle on the sides
lie on the same straight line.
(ii.) If four intersecting straight lines be taken three
together, so as to form four triangles, the orthocentres of
these trianorles lie on a rifjht line.
17. Describe a parabola through four given points.
18. A parabola rolls on an equal parabola, the vertices
originally coinciding. Prove that the tangent at the
vertex of the rolling parabola always touches a fixed
circle.
19. If two intersecting parabolas have a common focus,
the angle between their axes is equal to that which their
common tangent subtends at the focus.
20. AP, AQ, are two fixed straight lines, and B a fixed
point. Circles described through A and B cut the fixed
lines in P and Q. Prove that PQ always touches a
parabola with its focus at B.
CHAPTER II.
THE ELLIPSE,
DESCRIPTION OF THE CURVE.
Proposition I.
Given the focus, directrix, and eccentricity of an ellipse
to determine any number of points on it.
Let S be the focus, MXM' the directrix, and e the
eccentricity.
Through /S draw SX perpendicular to the directrix.
Divide SX in A, so that
SA=eAX.
60
ELLIPSE. 61
Also in XS' produced, take A' so that
SA' = eA'X*
Then A and A' are points on the ellipse and are its
vertices.
Take any point JV on A A'; through -ST draw PXP' per-
pendicular to AA'; with centre S and radius equal to
e . XX, describe a circle, cutting PXP' in P and P'. Then
P and P' shall be points on the ellipse. Draw PM, P'M^
perpendicular to the directrix.
Then SP = e. XX [Const
= e.PM,
and 8^=6. XN
= e.P'M'.
Therefore P and P' are points on the ellipse.
In like manner, by taking any other point on AA', any
number of points on the curve may be determined.
Def. The length of the axis intercepted between the
vertices (J. and J.') of the ellipse is called the major axis.
Del The middle point (C) of the major axis is called
the centre of the ellipse.
Def. The double ordinate {BCR) through the centre
(c) is called the minor axis of the ellipse.
Ex, 1. The ellipse is symmetrical ■with respect to its axis.
Corresponding to any point X on the line AA' we get two points
P and P', such that the chord PP" is bisected at right angles by the
axis A A'.
Ex. 2. Any two right lines drawn from any point on the axis to
the curve, on opposite sides of the axis and equally inclined to it,
are equal, and conversely,
Ex. 3. If two equal and similar ellipses have a common centre,
the points of intersection are at the extremities of central chorda at
right angles to each other.
* Since e is less than unity it is clear tliat A will lie between X
and S and A' without XS on the same side as S.
62 GEOMETEY OF CONICS.
Ex. 4. Prove that the ellipse lies entirely between the lines
drawn through A and A' at right angles to the axis.
In order that the circle may intersect PJVP' the point lY must be
so situated that aS'^V may not be greater than the radius of the circle
SF, that is, eJVX. It may easily be shown that this is the case only
when iV lies between A and A'.
Ex, 5. Show that as F moves from A to A', itsfocal distance
(SF) increases from SA to SA'.
For SF=e.^LY, and JVX has AX and A'X for its least and
greatest values respectively.
Ex. 6. Hence prove that the ellipse is a closed curve.
Ex. 7. If a parabola and an ellipse have the same focus and
directrix, the parabola lies entirely outside the ellipse.
Ex. 8. A chord QQ' of an ellipse meets the directrix in D.
Prove that
SQ : QD=SQ' : Q'l).
Ex. 9. A straight line meets the ellipse at F and the directrix
in J). From any point K in FB, K U is drawn parallel to DS to
meet SF in U, and KI is drawn perpendicular to the directrix.
Prove that SU=e. KI. (Cf. Prop. XVI., which is a particular case
of this.)
Ex. 10. A point F lies within, on or without the ellipse, accord-
ing as the ratio SF : FM is less than, equal to, or greater than the
eccentricity, FM being the perpendicidar on the directrix.
Proposition II.
The ellipse is symmetTncal with respect to the minor
axis and has a second focus (S') and directrix.
Let S be the given focus and MX the given directrix.
Take any point M on the directrix, and through the
vertices A and A' draw AH and A'H' at right angles to
AA', meeting the straight line through M and 8 at H and
//' respectively. Describe a circle on HH' as diameter
and through M draw MPP\ parallel to AA', to meet the
circle in P and P'. Then P and P' shall be points on the
ellipse.
For MH:HS=XA:AS=l:c,
and MH': B'S=XA': A'S= 1 : e.
J
ELLIPSK 53
Therefore MH : HS = MB' : H'S,
and the angle HPH' is a right angle. [Euc. III. 31.
Therefore, PH bisects the angle SPM.
Therefore SP:PJI = SH: HM
Therefore, P is a point on the ellipse. Similarly, it may-
be shown that P' is a point on the ellipse.
Again, the straight line drawn through 0, the centre of
the circle, at right angles to AA' will bisect both AA' and
PP" at right angles, and will therefore coincide with the
minor axis in position.
The ellipse is therefore symmetrical with respect to the
minor axis. [Def.
As the minor axis divides the curve into two parts such
that each is the exact reflexion of the other, if A'S' be
measured off equal to AS and A'X' = AX, and X'M be
54
GEOMETRY OF CONICS.
drawn at right angles to X'X, the curve could be equally
well described with S' as focus and X'lF as directrix.
The ellipse therefore has a second focus {S') and a
second directrix {X'M).
Ex. Every chord drawn through the centre (7 is bisected at tliat
point. (From the symmetry of the figure.)
From this property the point C is called the centre of the curve.
PROPEETIES OF CHORDS AND SEGMENTS OF
CHORDS.
Proposition III.
/71 tJie ellipse GA = e. OX (1)
08=6. CA (2)
CS.GX=CA'' (3)
We have, from the definition,
SA^e.AX,
SA' = e.A'X=e.AX\
1'herefore, by addition,
AA'=e{AX+AX')
= eXX\
Therefore GA=e.CX
.(1)
ELLIPSE.
oo
By subtraction, SS' = e(A'X - AX)
==e.AA'.
Therefore CS=e.CA (2)
Therefore CS . CX = CA'-. (n)
Ex. Given the ellipse and one focus, find the centre and the
eccentricity.
Describe a circle ■with S as centre, cutting the curve in P, f.
The axis bisects PF" at rij^ht angles.
Proposition IY.
The sum of the focal distances of any point on an
ellipse is constant and equal to the major axis.
Let P be any point on the ellipse. Join PS, FS', and
through P draw MP^F perpendicular to the directrices.
Then SP = e.PM,
ST = e.PM\
Therefore SP + S'P = e. (P1I/+ P^^)
= e.MM'
==eXX'
= AA'. [Prop. in.
Ex. 1. Show how to construct the ellipse mechanically.
First Method. — Fasten the ends of a string to two drawing pins
fixed at iS' and .S* on a board, and trace a curve on the board with
a pencil pressed against the string, so as to keep it always
56 GEOMETRY OF CONICS.
stretched. The curve traced out will be an ellipse, with foci at
S and S', and major axis equal to the length of the string.
Secotid Method. — Suppose two equal thin circular discs A and B,
attached to each other, to rotate in opposite directions round an
axis through their common centre ; and, suppose one end of a
fine string (which is wrapped round the discs, and passing through
small rings at C and D in the plane of the discs, is kept stretched
by the point of a pencil at P) to be wound on to its disc, while
the other is wound off. The curve traced by P will have the
pro]5erty CP + J)P= constant,
and will, therefore, be an ellipse.
Ex. 2. The sum of the focal distances of any point is greater
than, equal to, or less than the major axis, according as the point
is without, u])on, or within the ellipse, and conversely.
Ex. 3. The distance of either extremity of the minor axis from
either focus is equal to the semi-axis-major.
Ex. 4. A circle is drawn entirely within another circle. Prove
that the locus of a point equidistant from the circumferences of
the two circles, is an ellipse. [The centres will be the foci.]
Ex. 5. Two ellipses have a common focus, and their major axes
equal. Show that they cannot intersect in more than two points.
The conmion points may be shown to lie on the line bisecting at
right angles the line joining the second foci.
Ex. 6. Prove that the external bisector of the angle SPS'
cannot meet the ellipse again, and is, therefore, the tangent to the
ellipse at P, according to Euclid's conception of a tangent, (("f.
Prop. XVII.)
Prove also that every other line through P will meet the curve
again. [Apply Ex. 2.]
Ex. Y. The major axis is the longest chord that can be drawn
in the ellipse.
Joining the foci with the extremities of any chord, it may be
shown that twice the chord is less than the sum of the four focal
distances, that is, less than twice the major axis.
Ex. 8. In what position of P is the angle SP^ greatest?
[When 7' is at either extremity of the minor axis.]
Ex. 9. If r and R be the radii of the circles inscribed in and
described about the triangle SP,S', prove that Rr varies as SP.S'P.
Proposition V.
In the ellipse
ELLIPSK 57
Let J5 be an extremity of the minor axis. Join BS,
BS'.
Then SB+S'B=AA'. [Prop. IV.
But 8B=8'B.
Therefore SB = CA.
Therefore CB^ = SB'- - CS"- [Euc. I. 47.
= CA^-CS^
= SA.S'A. [EuaIL5.
Ex. 1 . Prove that e- = l — ..
Ex. 2. Prove that S'.S- = A' A- - SET-.
Ex. 3. If the angle SBS be a right angle, show that
CA^^i.CB.
Ex. 4. A circle is described passing through B and touching
the major axis in H ; if ><K be its diameter, prove that
SK.BC=AC-.
Ex. 5. Circles are described on the major and minor axes a.s
diameters. PP' is a chord of the outer circle cutting the inner in
q, q. Prove that pq . P'q=CS\
Ex. 6. Given a focus S and a point P on an ellipse, and the
lengths of the major and minor axes, find the centre.
On SP produced, take >S'A'' equal to the major axis ; *S" lies on the
circle with centre P and radius PK. On >S'A'a.s diameter describe a
circle, and place in it KIT equal to the minor axis ; S' lies on the
circle with centre *S' and radius SU.
58 GEOMETRY OF CONICS.
* Proposition VI.
The latus rectum of an ellipse is a third proportional
to the major and minor axes (SL = CB^/GA).
Let LSL' be the latus rectum. Draw LM perpendicular
t<:) the directrix.
Then
CS=eCA,
[Prop. III.
SL = eLM
[Def.
= eSX;
therefore
SL.CA = C8.SX
= CS{CX-CS)
= CS.CX-CS^
= CA^-GS-'
[Prop. III.
= C]?
[Prop. V.
Ex. 1. Construct on the minor axis as base a rectangle which
shall be to the triangle SLS' in the duplicate ratio of the major axis
to the minor axis.
Draw BK parallel to LS', meeting the major axis in A' ; the other
side of the rectangle = :^CA'.
Ex. 2. The extremities of the latera recta of all ellipses which
have a common major axis, lie on two parabolas.
If LN be perpendicular to CB, LN'^ = AC{AC- CN) ; hence, L lies
on a parabola of which CB is the axis, and the vertex is at a
distance from C= CA
ELLIPSE. 59
♦Proposition VII.
Any focal chord of an ellipse is divided liarmonicaU'if
hy the fociis and the directrix.
Produce the focal chord PSp to meet the directrix iu
D, and draw PM, pm perpendicular to the directrix.
Then PD:j)D = PM :i^m,
but PS = e.PM,
and pS=e.pm ;
therefore PD : pD = PS : pS.
Hence Pp is divided harmonically in S and D.
Ex. 1. Tlie semi-latus rectum is a harmonic mean between the
segments of any focal chord.
Ex. 2. Focal chords are to one another as the rectangles con-
tained by their segments.
Proposition VIII.
// any choixl QQ' of an ellipse intersects the directrix
in D, SD bisects the exterior angle bettveen SQ and SQ'.
Draw QM, QM' perpendiculars on the directrix, and
produce QS to meet the ellipse in q.
60
GEOMETRY OF CONICS.
Then, by similar triangles,
QD:Q'D = QM:Q'M'
=SQ:SQ';
therefore 8D bisects the exterior angle Q'Sq. [Euc ^'I. A.
Ex. 1. PSj) is a focal chord. Prove that 21 F and A'p are eqiially
inclined to the axis.
Ex. 2. Given the focus and three points on an ellipse, find the
directrix and the axis.
Ex. 3. If P be any point on an ellipse, and PA, PA' when
produced meet the directrix in J^J and F, show that £F subtends a
right angle at the foc;us.
Ex. 4. If A'S' be measured off along A'A equal to AS, and
A'X' be measured off along AA' equal to AX, and if PA and PA'
when pi'oduced meet the straight line through A'' at right angles to
the axis in E', F', show that E'X' . F'X' = EX. FX, and that E'F'
subtends a right angle at S'. (This is to be proved without assum-
ing the existence of the second focus and directrix of the curs-e.)
Ex. 5. Hence, show that if PK be tiie perpendicular on E'F',
S'P=e. PK ; and deduce the existence of a second focus and
directrix corres])onding to the vertex A'.
Ex. 6. If two fixed points Q, Q' on an ellipse be joined Avith a
third variable point 0 on the curve, the segment qq' intercepted on
either directrix by the chords QO and (^0 produced, subtends a
constant angle at the corresponding focus.
The angle qSq' may be proved to be equal to half of the anglo
QSq.
ELLIPSE. CI
Ex. 7. PSp is a focal chord ; 0 is any point on the curve ; PO,
pO produced meet the directrix in /), d. PS:ove that Dd subtends a
right angle at the focus.
Ex. 8. Given the focus of au ellipse and two points on the curve,
prove that the directrix will pass through a fixed point.
Ex. 9. A straight line which meets au ellipse will, in general,
meet it in two points, and no straight line can meet it in more
points than two.
The first part follows at once from the fact that the eUipse is a
closed curve. (Prop. I., Ex. 6. Cf. also Ch. I., Prop. X-, Ex. 8.)
Then, if the line meets the cur\-e in Q and Q', and the directrix in
2), SQ and S(/ will be equally inclined to DIS. Hence, if there be a
third point of intersection (jf', SQ' and »S'^ will make the same angle
with DS, which is impossible.
Proposition IX.
Tlie square of the ordinate of any point on an ellipse
varies as the rectangle under the segments of the axis
made hy the ordinate (PN^ : AX A'N=CB^ : CA"-).
Let PX be the ordinate of any point P on the ellipse.
Let PA and A'P produced meet the directrix in D and
jy. Join SD, SU, and SP, and produce PS to meet the
curve in p.
02 GEOMETRY OF CONICS.
Then, from the similar triangles PAN and DAX,
PN:AN=DX:AX.
Also, from the similar triangles PA'N and D'A'X^
PN.A'N=D'X:A'X',
therefore PN^ : AN . A'N= DX.D'XiAX. A'X.
Again, SD and 8D' bisect the angles pSX and PSX
respectively ; [Prop, VIII.
therefore the angle DSD' is a right angle, and
DX.D'X = SX^; [Euc.VI.8.
therefore PN'^ : AN . A'N= SX^-.AX. A'X.
But the ratio SX'-:AX .A'X is constant; therefore the
ratio PN^-.AN.A'N has the same value for all positions
of P.
In the particular case when P coincides with the
extremity B of the minor axis, the ratio PN^iAN.A'N
becomes GB^ : GA^ ; therefore
PN^ : AN. A'N= GB^ : GA\
P being any point on the ellipse.
Ex. 1. Prove that P^^^ : CA^ - CN^ = CB^ ; CA*.
Ex.2. Prove that ^+^^^- = 1,
Ex. 3. Prove that CP^=CB^+^. CJP ; and hence deduce that
of all lines drawn from the centre to the curve CA is the greatest
and CB the least. (See Prop. V., Ex. 1.)
Ex. 4. Show that FN increases as ^ moves from yl to C.
Ex. 5. If PM he drawn perpendicular to the minor axis, de-
duce that PM' : BM. B'M=^ CA^ : CB^.
Ex. 6. P, Q are two points on an ellipse. AQ, A'Q cut PA^ in
L and M respectively. Prove that PJV^ ^ xiV. MK
Ex. 7. Deduce Prop. VI.
Ex. 8. If JVQ be drawn parallel to AB, meeting the minor axis
in Q, show that PJV^ = BQ.BQ.
Ex. 9. If a point P moves such that PJV^ -.AN.A'N in a constant
ratio, PN being the distance of P from the line joining two fixed
ELLIPSE.
63
points A, A', and JT being bet-ween A and A', the locus of P is an
ellipse of which AA' is an axis.
Ex. 10. The locus of the intersection of lines dra-^Ti through
A, A' at right angles to JP, A'F, is an ellipse. [AA' will be the
minor axis. See Ex. 5, 9.]
Ex. 11. The tangent at any point P of a circle meets the
tangent at the extremity .4 of a fixed diameter AB in T. Find
the locus of the point of intersection (Q) of AF and BT.
QM being perpendicular to AB, the triangles QJIA, APB, and
ATG are similar ; so are the triangles QMB and TAB. Hence
QM-^:AM.BM=AC:AB.
Ex 12. The ordinat^s of all points on an ellipse being produced
in the same ratio, the locus of their extremities is another ellipse.
Ex. 13. P is any point on an ellipse ; AQO is drawn parallel
to CP meeting the curve in Q and CB produced in 0. Prove that
A0.AQ=2CB^.
Proposition X.
The loeiis of the middle jxnnts of any system ofjparallel
chords of an ellipse is a straight line passing through the
centre.
Let QQ' be one of a system of parallel chords and Fits
middle point.
Draw QM, Q'M perpendicular to the directrix. Draw
64 GEOMETRY OF CONICS.
SY perpendicular to QQ' and produce it to meet the
directrix in K. Produce QQ' to meet the directrix in
R. Join SQ, SQ\
Then SQ:SQ' = QM- Q'M'
= QR:Q'R
Therefore SQ' - SQ''' : QR^ - Q'R' = SQ^ : QR^.
But SQ^-SQ'^ = QY^--Q'l^
= (QY+Q'YKQY-Q'Y)
= 2QQ\YV.
Similarly QR^ - Q'R^ = 2QQ' .RV,
Therefore YV: R V= SQ'- : QRK
Now the ratio SQ : QM is constant, also the ratio
QM\ QR is constant, since QQ' is drawn in a fixed
direction. Therefore SQ : QR is a constant ratio.
Therefore also YV:RV is a constant ratio for all
chords of the system.
But as R always lies on a fixed straight line (the
directrix) and F on another fixed straight line (the
focal perpendicular on the parallel chords) intersecting
the former in K, V must also lie on a third fixed
straight line passing through the same point K.
Also C, the centre of the ellipse, is evidently a point
on this line, since the parallel chord through G is, from
the symmetry of the figure, bisected at that point.
Hence, the diameter bisecting any system of parallel
chords of an ellipse is a chord passing through its centre.
Ex. The diameter bisecting an}' system of parallel chords, meets
the directrix on the focal perpendicular ou the chords.
JVote.—See Prop. XI., Ex. 10.
Def. The circle described on the major axis (AA') as
diameter is called the auxiliary circle.
ELLIPSE. 66
Proposition XI.
Ordinates drawn from the same point on the axis to
the ellipse and the auxiliary circle are in the ratio of the
minor to the m.ajor aocis.
^^^^^HV
Let ApA' be the auxiliary circle and let NPp he a
common ordinate to the ellipse and the circle.
Then PN'^: AX. A'N= CB^ : GA\ [Prop. DL
and pN^ = AN .A'N. [EuaIIL3&35.
Therefore PN^ : pN~ = GB^ : CA\
Therefore PIf : pN= CB : GA.
Note. — By the help of this important property of the circle upoa
the major axis as diameter, many propositions concerning the
ellipse may be easily proved, as will be shown hereafter. Hence
the name auxiliary circle.
Defl The points P and p lying on a common ordinate
pPN of the ellipse and its auxiliary circle are called
corresponding 'points.
Ex. 1. A straight line cannot meet the ellipse in more than
two points. (Cf. Prop. YIII.. Ex. 9.)
6C GEOMETRY OF CONICS.
Ex. 2. PM drawn perpendicular to BB' meets the circle on the
lainor axis as diameter in p'. Prove that
PM'.p'M^CA'.CB.
(See Prop. IX., Ex. 5.)
Ex. 3. PNy PM are perpendiculars on the axes, meeting the
circles on the axes as diameters in p, p' respectively.
Prove that p and p' being properly selected, pp' passes through
the centre.
Ex. 4. Through P, KPL is drawn making the same angle with
the axes as pC, and cutting them in K and L. Show that KL is
of constant length. {KL = CA + CB.)
Ex. 5. If the two extremities of a straight line move along two
fixed straight lines at right angles to each other, any given point
on the moving line describes an ellipse.
Let the fixed straight lines intersect in 0, and let P be the given
point on the moving line AB oi which C is the middle point. Let
QPN drawn at right angles to OB, meet OC, OB in Q and N
respectively. Then, since OQ=AP, the locus of Q is a circle;
also, as PN:QN=PB:FA, the locus of P is an ellipse.
Ex. 6. Given the semi-axes in magnitude and position, construct
the curve mechanically.
Mark off on the straight edge of a slip of paper two lengths PA
and PB in the same direojtion and equal to the semi-axes respec-
tively. If the paper be now made to move so that A and B may
always be on the lines representing the axes in position, P will
trace out the ellipse. (See Ex. 5.)
Ex. 7. If a circle roll within another circle of double its radius,
any point in the area of the rolling circle traces out an ellipse.
First Method. — Let C be the centre of the rolling circle, and 0
that of the other. If the given point P be on the radius CM, M
will describe the diameter A'OA of the outer circle. Draw RPN
perpendicular to OA', meeting OC in R and OM in N. Then since
Ci2=CP, the locus of iil is a circle; and, as PN •.RN=PM :0R,
the locus of P is an ellipse.
Second Method. — The point J/ coincided with A' at the beginning
of the motion ; if in any position, the circles touch at §,
axe MQ=&rc A'Q, angle QCM=2 angle QOMy
.■. OCQ is always a straight line, so also is MCN, N being the
intersection of the inner circle with that radius of the outer which
is at right angles to OA. It is clear, therefore, that the motion of
a point P in J/]A^ is exactly the same as that of a point in the
moving rod in Ex. 6.
ELLIPSE 67
Ex. 8. From the centre of two concentric circles, a straight line
is drawn to cut tliem in P and Q ; through P and Q straight lines
are drawn parallel to two given lines at right angles. Prove that
the locus of their point of intersection is an ellipse, of which the
outer circle is the auxiliary circle.
Ex. 9. yPp, yPp' are ordinates of the ellipse and its auxiliary
circle. Show that PP, pj/ produced meet on the axis in the same
point T.
Ex. 10. Deduce from Ex. 9 a proof of Prop. X.
Let Vy vhe the middle points of PP, pp'. Vv produced bisects
yy at right angles in M. Now as long as PP remains parallel
to itself, pp' must remain parallel to itself, and, therefore, ita
middle point v lies on a fixed straight line, the diameter at right
angles to pp'. T, therefore, lies on a fixed straight line through
C, since vM : VM= CB : CA.
♦Propositiok XII.
If a system, of chords of an ellipse he drawn thi'ough a
fixed point the rectangles contained by their segments are
as the squares of the parallel serai-diameters.
Let QOQ' be one of the system of chords drawn through
the fixed point 0 and CP the semi-diameter parallel to
QQ'. Then QO.OQ: CF^ shall be a constant ratia
68 GEOMETRY OF CONICS.
Describe the auxiliary circle, and let p, q, q' be the cor-
responding points to P, Q, Q\ Join Cp and qq' and draw
through 0 a line perpendicular to the major axis, meeting
it in D and qq' in o.
Then, since QM : qM= QM : q'M
= GB:GA, [Prop. XL
the straight lines QQ' and qq' produced meet the axis pro-
duced in the same point T.
Again, the triangles FNG and QMT being similar
NGiMT=PN:QM
=pN:qM. [Prop. XL
Therefore the triangles pNG and qMT are similar.
[Euc. VI. 6.
Therefore pG is parallel to qT.
Therefore the triangles pPG and qQT are also similar.
Now QO:qo = QT:qT,
also OQ':oq' = QT:qT.
Therefore QO.OQ':qo.oq' = QT^ :qT^
= GI^:G^,
or QO . OQ' : GP' = qo . oq': GP\
Now, since OD:oD = GB:GA,
and the point 0 is fixed, the point o is also fixed ; hence
qo . oq' is constant. [Eua III. 35.
Also Gp = GA = constant.
Therefore QO.OQ'iGP^
is a constant ratio.
Ex. 1. The ratio of the rectangles under the segments of any
two intersecting chords of an ellipse, is equal to that of the rect-
angles under the segments of any other two chord:) parallel to the
former, each to each.
ELLIPSE. 69
Ex. 2. If two chords of an ellipse intersect, the rectangles
under their segments are as the parallel focal chords. (Apply
Prop. VIL, Ex. 2.)
Ex. 3. Ordinates to any diameter at equal distances from the
centre are equal,
Ex. 4. QCq is the central chord parallel to the focal chord
PSp. Prove that
SP.Sp:CQ.Cg=CJP: CA*.
* Proposition XIII.
If a circle intersect an ellipse in four points their
common chm'ds will he equally inclined, tvx> and two, to
the axis.
Let Q, Q\ q, q', be the four points of intersection.
Join QQ', qq\ intersecting in 0.
Then QO . OQ' ^ qO . Oq\ [Euc. Ill 35.
Therefore the semi-diameters parallel to QQ' and qq'
respectively, are equal to each other, [Prop. XIL
and they are, therefore, equally inclined to the axis from
the symmetry of the figure. (See also Prop. I., Ex. 2.)
Therefore, the chords QQ^ and qq' are equally inclined to
the axis.
70 GEOMETEY OF CONICS.
In like manner it may be shown that the chords Qq
and Q'q' as well as the chords Qq' and qQ' are equally
inclined to the axis.
Ex. 1. If two chords, not parallel, be equally inclined to the
axis of an ellipse, their extremities lie on a circle.
Ex. 2. If P be a fixed point on an ellipse and QQ' any ordinate
to CF, show that the circle QPQ' will intersect the curve in
another fixed point.
PEOPERTIES OF TANGENTS.
It has been already observed in Chapter I. that, gene-
rally, from a chord property of a conic a corresponding
tangent property may be deduced. The student should
work out the following exercises as illustrating the
method in the case of the ellipse.
* Deduce from Prop. XII. : —
Ex. 1. The tangents to an ellipse from an external point are
proportional to the parallel semi-diametei"s.
Ex. 2. If the tangents at three points F, Q, R on an ellipse,
intersect in r, q, p, show that
Fr.pQ.qR=Pq.rQ.pR.
Ex. 3. If two parallel tangents OP, O'P be met by any third
tangent OQCf, then OP . 0'P' = OQ . O'Q.
Ex. 4. If from any point without an ellipse a secant and also
a tangent be drawn, the rectangle under the whole secant and the
external segment is to the square of the tangent as the squares
of the parallel semi-diameters.
Ex. 5. If two tangents be drawn to an ellipse, any line drawn
parallel to either will be cut in geometric progression by the other
tangent, the curve and the chord of contact.
Ex. 6. Any two intersecting tangents to an ellipse are to one
another in the sub-duplicate ratio of the parallel focal chords.
Ex. 7. If two parallel tangents J Q and OR be cut by any third
tangent APO, and RP meets QA in B, show that AQ=AB.
♦Deduce from Prop. XIII. : —
Ex. 1. PQ, PQ" are chords of an ellipse equally inclined to the
axis. Prove that the circle PQQ' touches the ellipse at P.
ELLIPSE.
71
Ex. 2. PF is a chord of an ellipse parallel to the major axis ;
FQ, FQ are chords equally inclined to that axis. Sho-w that
Q^ is parallel to the tangent at P.
Ex. 3. If a circle touch an ellipse at the points P and §, proye
that PQ is parallel to one of the axes.
See also Props. XTV. and XV.
Proposition XTV.
The tangent to an ellipse at either end of a diameter is
parallel to the system of chords bisected by the diameter.
Let PVGP' be the diameter bisecting a system of
chords parallel to QQ'. Let QQ^ be made to move
parallel to itself so that Q may coincide with V. Since
QV is always equal to Q'F, [Prop. X.
it is clear that Q' will also coincide with V, and the
chord in this its limiting position will be the tangent
to the ellipse at P.
Ex 1. The tangent at the vertex is at right angles to the major
axis. [From symmetn', the chords at right angles to the major
axis are bisected by it.]
Ex. 2. The line joining the points of contact of two parallel
tangents is a diameter.
Ex. 3. Any tangent is cut harmonically by two parallel tan-
gents and the diameter passing through their points of contact.
(See note on Tangent Properties, I., Ex. 3.)
72 GEOMETRY OF CONICS.
Ex. 4. An ellipse is described about the triangle ABC, having
its centre at the point of intersection 0 of the medians. OA^ OB,
OC produced meet the ellipse in a, (i, y. Prove that the tangents
** «> A y ^o™i a triangle similar to ABC and four times as large.
Proposition XV.
The portion of the tangent to an ellipse at any point
intercepted betiveen that point and the directrix suhtends
a right angle at the focus, and conversely.
Also the tangents at the ends of a focal chord intersect
on the directrix.
First. — Let any chord QQ' of the ellipse intersect the
directrix in Z.
Then SZ bisects the exterior angle Q'Sq. [Prop. VIII.
Now, let the chord QQ' be made to turn about Q until
the point Q' moves up to and coincides with Q, so that the
chord becomes the tangent to the ellipse at Q. In this
limiting position of the chord QQ', since Q and Q' coincide,
the angle QSQ' vanishes and therefore the angle Q'Sq
becomes equal to two right angles. But, since SZ always
bisects the angle Q'Sq, in this case the angle QSZ is a
right angle.
ELLIPSE. 73
Again, let QZ subtend a right angle at S; then it shall
be the tangent to the ellipse at Q. For, if not, and if
possible, let QZ be the tangent at Q; then the angle
QSZ is a right angle, which is impossible. Therefore QZ
is the tangent at Q.
Secondly. — Let QSq be a focal chord and QZ the tan-
gent at Q. Join ZS, Zq.
Then the angle QSZ being a right angle, the angle
ZSq is also a right angle, and therefore qZ is the tangent
to the ellipse at q. Therefore the tangents at Q and q
intersect on the directrix.
Ex. 1. Tangents at the extremities of the latus rectum inter-
sect in X
Ex. 2. If through any point F of an ellipse, an ordinate QPN
be drawn, meeting the tangent at L in Q, prove that Qy=SP.
Ex. 3. To draw the tangent at a given point P of an ellipse.
Ex. 4. By drawing the tangent at B, prove that CS. CX=CA*.
^ Ex. 5. If ZQ meets the other directrix in Z'y ZP subtends a
right angle at S.
Ex. 6. If QZ intersect the latus rectum in 2), prove that
SB^e.fiZ.
74
GEOMETRY OF CONICS.
Proposition XVI.
If from a point 0 on the tangent at any point P of an
ellipse perpendiculars 0 U and 01 be drawn to SP and
the directrix respectively, then
SU=e.OI,
and conversely.
Join SZ and draw PM perpendicular to the directrix.
Because ZSP is a right angle, [Prop. XV.
ZSf is parallel to OU.
Therefore, by similar triangles,
SU:SP = ZO:ZP
= OI:PM.
But SP^e.PM;
therefore 811=6. 01.
Again, for the converse proposition, if a line OP meets
the ellipse at P, and the same construction is made as
before, we have
8U=e.0I,
and SP = e.PM;
therefore SU: SP=0I:PM
==Z0:ZP.
ELLIPSE 75
Therefore OU is pamllel to ZS, [Euc. VI. 2.
and the angle PSZ is a right angle.
OP is, therefore, the tangent at P. [Prop. XV.
Note. — See Chap. I., Prop. XTTL, also Prop. I., Ex. 9.
Proposition XYII.
The tangent at any ix^int of an ellipse makes eqiud
angles with the focal distances of the 'point.
Let the tangent at F meet the directrices in Z and Z'.
Draw MPM' perpendicular to the directrices, meeting
them in M and M' respectively. Join SP, SZ, S'P, and
S'Z'.
Then, in the two triangles PSZ and PS'Z', the angles
PSZ and PS'Z' are equal, being right angles, [Prop. XV.
and SP:S'P = PM:PM'
= PZ:PZ',
and the angles PZS and PZ'S' are both acute angles.
Therefore the triangles are similar ; [Eua VI. 7.
therefore the angle <SPZ=the angle S'PZ'.
76 GEOMETRY OF CONICS.
Ex. 1. If a line drawn tlirough P bisect the exterior angle
between SP and S'P, it will be the tangent at P.
Ex. 2. The tangent at the vertex is at right angles to the
major axis.
Ex. 3. The perpendiculars from Z and Z' on SP intercept a
length equal to AA'.
Ex. 4. The tangent at any point makes a greater angle with
the focal distance than with the perpendicular on the directrix.
Ex, 5. If SY, S'Y' be the perpendiculars upon the tangent at P,
and PiV be the ordinate of P, prove that jTiV bisects the angle
Ex. 6. If ST, the perpendicular on the tangent at P, meet S'P
produced in s, prove that
(i) sr=sr, (ii) sp=Ps, (iii) s's==aa'.
On account of property (i), s is called the image of the focus in the
tangent.
Ex. 7. Prove that the locus of the image of the focus in the
tangent is a circle.
The circle, of which the centre is a focus and the radius equal
to the major axis, is sometimes, though i^.ot quite properly, called
the Director Circle, by way of analogy to the directrix of the
parabola, which is, in the case of that curve, the locus of the image
of the focus in the tangent. (See Chap. I., Prop. XIV., Ex. 7.)
Ex. 8. Given a focus and the length of the major axis, describe
an ellipse touching a given straight line and passing through a
given point. (Apply Prop. IV. ; Newton, Book I., Prop. XVIII.)
Ex. 9. Given a focus and the length of the major axis, describe
an ellipse touching two given straight lines. (Apply Prop. IV.,
cf. Prop. XXIII., Ex. 4 ; Newton, Book I., Prop. XVIII.)
Ex. 10. If a circle be described through the foci of an ellipse,
a straight line drawn from its intersection with the minor axis
to its intersection with the ellipse, will touch the ellipse.
Proposition XVIII.
To draw two tangents to an ellipse from an external
point.
Let 0 be tTie external point. Draw 01 perpendicular
to the directrix, and with centre S and radius equal to
ELLIPSE. 77
e . 01, describe a circle. Draw OU, OTJ' tangents to this
circle, and let SU, SU' meet the ellipse in Q, Q\ Join
OQ, OQ'. Then OQ, OQf shall be the tangents required.
For OfT" is at right angles to SQ, [Euc. IIL la
and SU=e.OL
Therefore OQ is the tangent to the ellipse at Q.
[Prop. XVL
Similarly OQf is the tangent at Q'.
Ex. 1. Alternative CoTistructton. — With centre 0 and radios OS
describe a circle ; with centre S' and radius equal to the major
axis describe another circle intersecting the former in M and M'.
Join S'M and S'M', meeting the ellipse in Q and Q' ; OQ, OQ' are
the tangents required. [The angle 0^3/= the angle OQS. Then
apply Prop. XVII., Ex. 1. It may be shown that the construction
given in Chap. I., Prop. XVI., is immediately deducible from this.]
Ex. 2. Show that only two tangents can be drawn to an ellipse
from an external point. (See Note to Chap. I., Prop. XVI.)
Proposition XIX.
The two tangents which can he drawn to an ellipse
from an external point subtend equal angles at the
focus.
78 GEOMETEY OF CONICS.
Let OQ, OQ' be the two tangents from 0.
Join SO, SQ, SQ', and draw 01, OU, OU' perpendi-
culars upon the directrix, SQ, SQ' respectively.
Then SU==e. OI=SU\ [Prop. XVI.
Therefore OU=OU'. [Euc. I. 47.
Therefore the angles OSU and OSU' are equal,
[Euc. I. 8.
and they are the angles which the tangents subtend at
the focus S.
Ex. 1. QQ produced meets the directrix in Z. Prove that OZ
subtends a right angle at S. [Prop. XV. is a particular case of
this.]
Ex. 2. If P be any point on an ellipse, the centre of the circle
touching the major axis, SP, and S'P produced lies on the tangent
at the vertex.
Ex. 3. The two foci and the intersections of any tangent with
the tangents at the vertices, are concyclic points.
Ex. 4. A variable tangent meets a fixed tangent in T. Find
the locus of the intersection with the variable tangent of the straight
line through S at right angles to ST.
[The locus is the tangent at the other extremity of the focal chord
through the point of contact of the fixed tangent.]
Ex, 6. The tangents at the ends of a focal chord meet the
I
ELLIPSE. 79
tangents at the vertex in T^ and T^. Prove that ATi.AT^ is con-
stant. (=-AS\)
Ex. 6. The angle subtended at either focus by the segm^t inter-
cepted on a variable tangent by two fixed tangents is constant.
Ex. 7. If OS intersect QQ^ in R and JtK be drawn perpen-
dicular to the directrix, prove that QK, Q'K are equally inclined
to the axis.
Ex. 8. An ellipse is inscribed in a triangle ; if one focus moves
along the arc of a circle passing through two of the angular points
of the triangle, find the locus of the other focus. [An arc of a circle
through the same angular points.]
Ex. 9. If a quadrilateral circumscribes an ellipse, the angles
subtended by opposite sides at one of the foci are together equal to
two right angles.
♦Proposition XX.
The two tangents drawn to an ellipse from, an external
point are equally inclined to the focal distances of that
point.
Let OQ, OQ be the two tangents from 0.
Join SQ, SO, SQ', S'Q, S'O, S'Q\ and produce SQ to R.
Let H be the point of intersection of SQ' and S'Q.
80 GEOMETRY OF CONICS.
Then
the angle SOQ= the angle OQR- the angle 08Q
[Euc. I. 32.
= half the angle /Sf'QiJ- half the angle QSQ'
[Props. XVII. and XIX.
= half the angle /SfHQ.
Similarly,
the angle ^'0Q' = half the angle S'HQ'.
Therefore,
the angle SOQ= the angle S'OQ'. [Euc. I. 15.
Ex. 1. Given two tangents to an ellipse and one focus, show
that the locus of the centre is a right line.
Ex. 2. On OQ, OQ' take OK, OK' equal to OS, OS' respectively.
Prove that KK' is equal to the major axis. [If SQ produced to
E be equal to the major axis, the triangles SOE and KOK' are
equal.]
Ex. 3. The straight line joining the feet of the perpendiculars
from a focus on two tangents is at right angles to the line joining
the intersection of the tangents with the other focus.
Ex. 4. The exterior angle between any two tangents is half the
sum of the angles which the chord of contact subtends at the foci.
[Cf. Chap. I., Prop. XIX.]
Ex. 5. The angle between the tangents at the extremities of a
focal chord is half the supplement of the angle which the chord
subtends at the other focus.
Ex. 6. Prove that
lS0S'+lS'Q0 + lSQ'0 = 2 right angles.
Ex. 7. If from any point on an ellipse tangents are drawn to
a confocal ellipse, these tangents are equally inclined to the tangent
at that point.
Def. Ellipses which have the same foci are called confocal
ellipses.
Ex. 8. If a perfectly elastic billiard ball lies on an elliptic
billiard table, and is projected in any direction along the table,
show that the lines in which it moves after each successive impact
touch a confocal conic.
Ex. 9. Normals at the extremities of a focal chord intersect in €>,
and the corresponding tangents meet in T. Prove that OT passes
through the other focus.
ELLIPSR
81
Proposition XXI.
The tangents at the extremities of any cJtorcl of an
ellipse intersect on the diameter which bisects the choi'd.
Let QQ'be the chord,and qq any other chord parallel to it.
Let qQ and q'Q' produced meet in 0. Bisect QQ in
Tand let OF meet 35^ in v.
Then QV.qv^OV.Ov
= QfV:g^v.
But QV=QfY.
Therefore qv = c^v.
Therefore OVv is the diameter bisecting the system of
chords parallel to QQ. [Prop. x.
82 GEOMETRY OF CONICS.
If now the chord qcl be made to move parallel to itself
until it coincides with QQ, qQO and q'Q'O will become
the tangents to the curve at Q and Q' respectively, and
they thus meet on the diameter bisecting QQ'.
Ex. 1, The diameter of an ellipse through an external point
bisects the chord of contact of the tangents from that point.
Ex. 2. Given a diameter of an ellipse, to draw the system of
chords bisected by it.
Ex. 3. The tangent at any point P of an ellipse meets the
tangent at A in Y. Prove that CF is parallel to A' P.
Ex. 4. If OPCP' be a diameter through 0, OQ a tangent from
0, and Q V be drawn parallel to the tangent at P, then
OP.OP'=OG.OV.
Hence show that OP : OP' = PV: P' V. [This shows that PP' is
divided harmonically in T'^and 0.]
Ex. 6. If any line drawn parallel to the chord of contact of two
tangents to an ellipse meets the curve, the segments intercepted
between the curve and the tangents are equal.
Proposition XXII.
If the tangent at any point Q of an ellipse meets any
diameter GP produced in T, and if QVhe the ordinate
to that diameter,
CV.CT=CI^.
Draw the tangent PR at P, meeting QT in B, and
draw PO parallel to QT meeting QFin 0.
ELLIPSE. 83
Then since POQR is a pai-allelogiam, RO bisects PQ,
and therefore passes through the centre C.
[Prop. XIV. and XXL
By similar triangles
CV: CP = CO:GR = CP:CT.
Therefore CV. CT= CP^.
Xote. — When the diameter coincides with the major
axis, the result is stated thus : —
Iftlie tangent at Q meets the major axis produced in
T, and QX be the 'perpendicular on the major axis,
CN.CT=CA\
When the diameter coincides with the minor axis, the
result is stated thus : —
If the tangent ai, Q meets tJie minor axis produced in
t, and Qn he the perpendicular on the minor axis,
Cn.Ct^CW.
These two particular cases are important, and should be carefully
noted by the student.
Ex. 1. VH drawn parallel to PQ meets CQ in. R. Prove that
PH is pai-allel to the tangent at Q.
Ex. 2. If a series of ellipses have the same major axis, the tan-
gents at the extremities of theii* latera recta meet at the same point
on the minor axis.
Ex. 3. If PT be a tangent to an ellipse meeting the axis in Tj
and AP, A'P be produced to meet the perpendicular to the major
axis through 7' in ^ and Q', then QT=(j[T. [If PXhe the ordinate
of P, the relation CT : C'A = CA : CX gives A'T : A'y=AT: AX.^
Ex. 4. If PX be perpendicular to the major axis, and the tangent
at P meet the major axis produced in T, any circle through ^V and
T cuts the auxiliarv circle at right angles. [If E be the centre of
tho circle, show that EX^ + CA'=EC'-.'\
Ex. 5. The locus of the middle points of all focal chords of an
ellipse is a similar ellipse.
Let 0 be the middle point of a focal chord PSp, and let the
tangent at Q where CO produced meets the curve, meet the major
84 GEOMETEY OF CONICS.
axis in T. If OM and QN be the ordinates to the major axis, it
readily follows that
CM. ISM CN.TN AN.A'N'
Then apply Prop. IX., Ex. 9.
Ex. 6. If CY, AZhe the perpei«liculars from the centre and an
extremity of the major axis on the tangent at any point P, show
that CA.AZ=Cr.Aiy.
Ex. 7. If a variable tangent to an ellipse meet two fixed parallel
tangents, it will intercept segments on them whose rectangle is
constant.
Let the tangent at Q meet the two parallel tangents PR and pr
in R and r. Pp is a diameter (Prop. XIV., Ex. 2). Let CD be the
semi-diameter parallel to PR meeting Rr in t. Let ^T''and Qv be
ordinates to CP, CD ; and let rR, pP meet in T. Then apply the
proposition with respect to the diameters CD, CP.
Ex. 8. In Ex. 7 prove that the rectangle under the segments of
the variable tangent is equal to the square of the semi-diameter
drawn parallel to it. (See Note on Tangent- Properties Ex. 1, 2.
Newton, Book I., Lemma XXIV.)
Ex. 9. If P is any point on the ellipse, find the locus of the
centre of the circle inscribed in the triangle SPS', [An ellipse. If
ON be the perpendicular from the centre 0 on A A', it mav be
shown that ON-^ : JS . NS' = aS-^I^ : CEK
Then apply Prop. IX., Ex. 9.]
Ex. 10. CD, CP are two semi-diameters of an ellipse. Tangents
at D and P meet CP and CD in K and T resiDcctively. Prove that
the triangles CDK and CPT aie equal in area.
Proposition XXIII.
The locus of the foot of the 'perpendicular drawn from
either focus upon any tangent to an ellipse is the
auxiliai'y circle ; and the rectangle under the focal
perpendiculars on the tangent is equal to the square of
the semi-axis minor. (SY.8'Y' = GB^.)
Let 8Y, S'Y be the focal perpendiculars upon the tan-
gent at any point P.
Join SF and ST. Produce S'F to meet SY in R.
Join GY.
ELLIPSE. 85
Then in the triangles SPY, RPY,
the angle 6'PF= the angle STT [Prop. XVIL
= the angle RPY, [Eue. I. 15.
and the angles SYP, RYP are equal, each being a right
angle, and YP is common,
therefore SP = PR.
and SY= YR. [Euc 1. 26.
Also SOLOS',
therefore CY is parallel to S'R, [Euc VI. 2.
therefore CY= | . S'R [Euc. VI. 4.
= h{S'P+PR)
= l(S'P-j-SP)
= hAA' [Prop. IV.
= 'CA.
Therefore the locus of Y is the auxiliary circle.
Similarly it may be shown that the locus of Y is the
same circle.
86 GEOMETRY OF CONICS.
Again, produce YG and Y'8' to meet in 2/, then y will
be on the auxiliary circle.
For, since (75=678' and 8Y is parallel to S'y the tri-
angles SGY and S'Gy are equal. [Euc. I. 26.
Therefore Gy = GY= GA, showing that y is on the auxiliary
circle.
Also SY=8'y
Therefore SY. ST^S'y . S'Y'
= S'A' . 8' A [Euc. III. 35.
= GB\ [Prop. V.
Ex. 1. CE parallel to the tangent at P meets SP, S'P in B, E'.
Prove that
(i) PE^PE' = CA.
(ii) SE=SE'.
(iii) the circle circumscribing the triangles CSE and CH'E' are
equal.
Ex. 2. The central perpendicular on the tangent at P meets SP
produced in Q, Prove that the locus of <? is a circle. [Centre B.
Radius = (Til.]
Ex. 3. If from the centre of an ellipse lines be drawn parallel
and perpendicular to the tangent at any point, they enclose a part
of one of the focal distances of that point equal to the other.
Ex. 4. Given a focus and the length of the major axis, describe
an ellipse touching two given straiglrt lines.
Ex. 5, Given a focus, a tangent, and the eccentricity, the locus
of the other focus is a circle. [Since C8=e.CY, the locus of the
centre is obviously a circle.]
Ex. 6. Prove that the perimeter of the quadrilateral SYY'S' is
the greatest possible when TV subtends a right angle at the centre.
Ex. 7. A line is drawn through >S" parallel to SP meeting YSin
0. Prove that the locus of l^is a circle.
Ex. 8. The right line drawn from either focus to the adjacent
point of intersection of any tangent with the auxiliary circle is per-
pendicular to the tangent.
Ex. 9. If through any point }' on the auxiliary circle YP be
drawn at right angles to SY, YP will be a tangent to the ellipse.
E.\. 10. If the vertex of a right angle moves on a fixed circle,
i
ELLIPSK 87
iiud one leg passes through a fixed point, the other leg will always
touch an ellipse. {Ci. Cliap. T., Prop. XXIII., Ex. 4.)
Ex. 11. Given the major axis and a tangent, show that the
directrix passes through a fixed point.
Ex. 12. The circle described on SP a.s diameter touches the
auxiliary circle.
Ex. 13. Given a focus, a tangent, and the length of the major
axis, the locus of the centre is a circle.
Ex. 14. Given the foci and a tangent, construct the ellipse
Ex. 15. Alternative Construction for Prop. XVIII.
Let 0 be the external point. On OS&a diameter describe a circle
intersecting the auxiliary circle in Fand Y' Then OF and OT'
{produced will be the tangents required.
Ex, 16. The right line drawn from the centre parallel to either
focal radius vector of any point on an ellipse to meet the tangent at
that point, is equal to the semi-axis major.
Ex. 17. Draw a tangent to an ellipse parallel to a given straight
line,
Ex. 18. Two ellipses, whose axes are equal, each to each, are
placed in the same plane, with their centres coincident and axes
inclined to each other. Draw their common tangents. [The com-
mon tangents pass through the points in which the lines joining
the foci of the curves meet the common auxiliary circle.]
Ex. 19. Given a focus, a tangent, and the length of the minor
axis, the locus of the other focus is a straight line.
Ex. 20. If the rectangle under the perpendiculars from the fixed
pomts on a right line be constant (=i'^), the line always touches
an ellipse of which the fixed points are the foci, and the minor
axis = 2h.
Ex. 21 A chord of a circle, centre C and radius r, subtends a
right angle at a fixed point 0. Prove that it always touches an
ellipse, of which C and 0 are the foci, and the square of the semi-
axis minor =r^ <^* CO-.
Ex. 22. If a second tangent to the ellipse intersect TPT' at
right angles in 0, prove that OY. OY' = CB^.
Hence, prove that CO- = CA^ -\- CD\
(a. Prop, XXIV.)
* Proposition XXIY.
The locus of Hie intersectuyn of tangents to an ellipse
which cut at right angles is a circle.
88 GEOMETRY OF CONICS.
Let the tangents OT, OT cut at right angles at 0.
Draw SY, OK perpendicular to OT and 8U, GK' per-
pendicular to OT. Join CY, GU, GO. Let GK, SU
intersect in H.
Now Y and U are on the auxiliary circle, [Prop. XXIII.
therefore GY=GU = GA.
Then GO"^ = GK'~ + GK"' [Euc. I. 47.
and CY^ = GK^+YK\
therefore GA^ = GK^ + SH' ;
also GU^ = GK'^+UK'\
therefore GA^ = GK'' + GE\
therefore 2GA'=^ GK'~ + GK'- +SIP+ HG^
= GO'' + GS^; [Euc. I. 47.
but GS^ = GA-'- GB\ [Prop. V.
therefore CO^ = GA'' + GB\
Hence the locus of 0 is a circle described with the centre
G and radius equal to AB.
Note. — This circle is called the Director Circle of the ellipse.
1
ELLIPSE. 89
Ex. 1. An ellipse slides between two fixed Hues at i-ight angles
to each other ; prove that the locus of its centre is an arc of a
circle.
Ex. 2. Any rectangle circumscribing an ellipse is inscribed in
the director circle.
Proposition XXV.
Tangents at coi^esponding 'points of an ellipse and its
auxiliary circle intersect on the major axis.
^
Let the ordinate pPN meet the ellipse in P and the
auxiliary circle in the corresponding point j^- Let qQM
be any other ordinate.
Then, because
QM:qM=CB:CA
= Pj^:pX, [Prop. XI.
the straight line QP, qp produced meet the major axis in
the same point T.
Now, if qQM be made to move parallel to itself so as to
coincide with j^P^, the points Q, P and q, p will coalesce,
and the chords QPT and qj^T will become tangents to the
ellipse and the circle at P and p respectively.
Ex. 1. Deduce this proposition from the property OK. CT= CA\
(Prop. XXII.)
Ex. 2. The tangent at p meets CB produced in Jv. Prove that
CK.PA'^CA.CB.
90 GEOMETRY OF CONICS.
Ex. 3. Show that the locu3 of the intersection of the normals
at P and jo is a circle of which the radius is CA + CB. [If the
normals intersect in 0, and if PR be drawn parallel to the major
axis to meet CO in 11, then, bv similar triangles, it may easily be
shown that Oli = CA, CIt = CB.]
Ex. 4. OQ, OQ' are tangents to an ellipse ; OJV is drawn per-
pendicular to the axis. Prove that the tangents to the auxiliary
circle at the corresponding points q, q' meet on 0^1
If QQ' produced meet the major axis in T, prove also that
Ciy.CT=C'A'.
[For the second i)art, note that if OJV meet the auxiliary circle
in R, the tangent at R meets the major axis at the point where
QQ', qq' meet it. Cf. also Prop. XXII., note, which is a limiting
case.]
Ex. 5. In Ex. 4, if OJV meets the ellipse in r, the tangent at r
intex'sects the major axis in 2\
PEOPERTIES OF NORMALS.
Proposition XXVI. ^
The nonnal at any point of an ellipse bisects the
angle between the focal distances of the point.
Let the normal at the point P meet the major axis
in G. Let YPY' be the tangent at P.
Then the angle SFY= the angle STY'. [Prop. XVIL
ELLIPSE. 91
But the angles GPY, GPY' are equal, being right angles ;
[Def.
therefore the angle SPG = the angle S'PG.
Ex. 1. If the tangent and normal at any point P meet the
muior axis in t and g, then P, t, g, S, and S' lie on the same circle.
Ex 2. Prove that the triangles SPG and gPS" are similar.
Ex. 3. If from g a perpendicular gK be drawn on SP or S'P,
ahowthAtPK=CA.
Ex.4. Prove that SP.SP=PG.Pg. [The triangles PSg,
PS'G are similar. Ex. 1.]
Ex. 5. No normal can pass through the centre, except it be at
an end of one of the axes.
Ex. 6. The normal PG and the focal perpendiculars on the
tangent at P are in harmonic progression.
Ex. 7. The circle described on PG as diameter cuts SP, /S'P in
K and Z. Prove that PG bisects KL at right angles.
♦Proposition XXVII.
If the normal at any point P of an ellipse meets the
major aads in G, SG = e . SP.
Join S'P.
Then, since PG bisects the angle SPS', [Prop. XXVI.
SG : S'G = SP : S'P ; [Euc VL 3.
therefore SG:SG+S'G = SP:SP+ S'P,
or SG : SP = SG -{-S'G: SP + S'P.
92
GEOMETRY OF CONICS.
But
and
therefore
SG-^S'0==8S' = e.AA\ [Prop. III.
HP-\-ST = AA'; [Prop. IV.
SG = e.SP.
Ex. 1. Show how to ilraw the iioriaal at any point without
drawing the tangent.
Ex. 2. If PM be drawn perjjendicular to the directrix, and
JfS meet the minor axis in g, sliow that Pg is the normal at P.
Ex. 3. A perpendicular is drawn from a fixed point Jlf on the
major axis of an ellipse, on the tangent at any point P. The locus
of the intersection of this perpendicular with SP is a circle.
Ex. 4. If GB be perpendicular to SP, prove that PE is equal
to half the latus rectum. [PS^ and SBG are similar triangles ;
therefore SE=e.SN, SP=e.NX,so that PE=e.SX.'\
Ex. 5. In Ex. 4, show that GE=e.PN.
Ex. 6. Show that
PG^:SP.S'P=CJ^'.CAK
(Cf. Prop. XXVI., Ex. 4, and Prop. XXVIII.)
Proposition XXVIII.
The normal at any point of an ellipse, terTninated hy
e'ltJier axis, varies inversely as the central perpendicular
on the tangent. (PG . PF= GB^ Pg . PF= GA^.)
Let the normal at P meet the major axis in G and thri]
minor axis in g ; let the taugent at P meet them in
ELLIPSE. 93
and t respectively. Draw PK, Pn perpendicular to the
major and minor axis, and let a straight line through the
centre, drawn parallel to the tangent at P, meet PJV, PG,
and Pn produced, in R, F, and r respectively.
Then, since the angles at X and F are right angles,
G, F, R, N lie on a circle ; therefore
PG.PF= PX . PR [Euc IIL 36.
= Cn . Ct [Euc. I. 34.
= CR^. [Prop. XXII., Xote.
Again, since the angles at n and F are right angles, g, F,
n, r lie on a circle ; therefore
Pg.PF=Pn.Pr [Euc. IIL 36.
= CN.CT [Euc. L 34.
= CA-. [Prop. XXTL, Note.
Therefore both PG and Pg vary inversely as PF, which
is equal to the central perpendicular upon the tangent
at P.
Ex. \. If CF meet the focal distances of i* in ^ and K, prove
that Pg subtends a right angle at E and E. (See Prop. XXIII.,
Ex. 1.)
Ex. 2. If the circle throiigh S, P, S' meets the minor axis iu
// on the side opposite to P, prove that .S^ varies as PG.
Ex. 3. PQ is drawn at right angles to SP, meeting the diameter
I>arallel to the tangent at P in Q. Prove tliat PQ varies inverselv
as PX.
Proposition XXIX.
If the Tiormal at any point P on an ellipse meets the
major aans in G, and PX he the ordinate to that axis,
(i) GN:CX=CBr-:GA\
(ii) CG = e\CX.
94- GEOMETRY OF CONICS.
Let the normal meet the minor axis in g. Draw Pn
perpendicular to the minor axis, and GF parallel to the
tangent at P.
SB^^B^^^I
Then, because the triangles PNG and Png are similar,
GN:CN=PG:Pg [Euc. VI. 2.
= PG.PF:Pg.PF
= CB^:CA^; [Prop. XXVIII.
therefore ON-GN:CN= GA^ - GB^ : GA\
or GG : GN= G8^ : GA\ [Prop. V.
But GS=e.GA; [Prop. III.
therefore GG = e^.GN.
Ex. 1. In the figure of Prop. XXVIII., prove that :—
(i) CG.CT=CS^.
(ii) Cg.Ct=CS\
(iii) J^G.CT=CI?.
(iv) Tg, tG intersect at right angles.
Ex, 2. Find a point P on the ellipse such that PO may bisect
the angle between PC and PN.
Ex. 3. In the figure of Prop. XXVI II., prove that the rect-'
angle under the focal perpendiculars on PG=CF. PT.
PROPERTIES OF CONJUGATE DIAMETERS.
Proposition XXX.
If one diameter of an ellipse bisects chords parallel to\
ELLIPSK 95
a second, the second diameter bisects citords lyarallel to
the first.
Let CP bisect chords parallel to CD ; then CD bisects
chords parallel to CP.
Draw A'Q parallel to CD, meeting CP in V; join AQ,
meeting CD in U.
Then A'Q is bisected in V and AA' in C\ therefore
CV Ls parallel to A Q. [Euc VI. 2.
Again, since AA' is bisected in (7, and CD is parallel to
^'Q, ^Q is bisected by CD. [Eua A'L 2.
Therefore CD bisects all chords parallel to J.Q, [Prop. X
and therefore all chords parallel to CP.
Def. Two diameters so related that each bisects chords
parallel to the other are called Conjugate Diameters.
Thus CP and CD are conjugate to each other ; so also are the
major and minor axes.
Ex. 1. If one diameter is conjugate to another, the first is
parallel to the tangent at an extremity of the second. (Prop. XIV.)
Ex. 2. Given an ellipse and two conjugate diameters, show how
to draw the tangent at any point.
If CP, CD be conjugate diameters, and QV is drawn parallel
to CD, QV is the ordinate to CP. In CP produced take T, such
that CV.CT=CP^. QT is the tangent at Q. (Prop. XXII.)
9C GEOMETRY OF CONICS.
Ex. 3. If CQ be conjugate to the normal at /", then CP is con-
jugate to the normal at Q.
Ex. 4. The focal perpendiculars upon CP and CD, when pro-
duced backwards, will intersect CD and CP on the directrix.
(Apply Prop. XXIX., Ex. 2.)
Ex. 5. The focus is the orthocentre of the triangle formed by
any two conjugate diameters and the directrix. (See Prop. X'.,
Ex. 1.)
Ex. 6. Any diameter is a mean j^roportional between tlie focal
chord parallel to it and the major axis. [The conjugate diameter
CD will bisect the focal chord. Then applv Prop. XXII., and
Prop. XXIII., Ex. 16.]
Ex. 7. The rectangle under the intercepts on any tangent
between the curve and any two conjugate diameters, is equal to
the square of the semi-diameter parallel to the tangent, and
conversely.
Let the tangent at Q meet the conjugate semi-diameters CP,
CD in T, T, and let CR be the semi-diameter parallel to TT'.
Let the tangent at R pai'allel to CQ meet CD in t. Draw the
ordinates Q V, Rv with respect to CD, parallel to CP. Then
C V. CT = Cv . Ct = CDl [Prop. XXII.
By similar triangles,
QT'.CR=CV:Cv = Ct:CT' = CR:QT'.
Therefore QT . QT' = CR\
Ex. 8. Given in magnitude and position any two conjugate
semi-diameters CP, CD of an ellipse, find the major and minor axes.
Produce CP to K, such that CP.PK=-CDK Bisect CK in 0,
and let the line tlu-ough 0 at right angles to CK meet the line
through P parallel to CD in //. With centre H and radius HC,
describe a circle cutting PH in T, T' ; the circle will also pas.s
through K. Then CT, CT' will coincide with the directions of
the major and minor axes respective! v.
For PT . PT' = CP . PK=CD- ; therefore CT, CT are conjugate
diameters (Ex. 7), and as they are at right angles, they must
coincide with the directions of the major and minor axes. (Cf.
Prop. XXXIII., Ex. 3 ; see also Miscellaneous Examples, 13, 14,
15, 16.)
To determine the magnitudes of the axes, observe that TPT' is
the tangent at P, and apply Prop. XXII., note.
Ex. 9. PP' is a fixed line. Find the locus of a point Q which so
moves that Q V being drawn in a fixed direction to meet PP' in V,
Q V^ is to PV . P' Fin a given ratio.
Bisect PP' in C, and through C draw CD in the fixed direction,
Buch that CD^ is to CP^ in the given ratio. Then thtf locus of Q
i
ELLIPSE.
97
will be the ellipse described with CP and CD as conjugate semi-
diameters (Ex. 8). Apply Prop. XII., and cf. Prop. XXXII.
Note.—liQV^-=PV. P' r, the semi-diameters CP, CD will be equi-
conjugate. In this case the position of the major and minor axes
may be at once determined, as they bisect the angles between the
equiconjugate diametei-s. (See Prop. XXXI., Ex. 3 )
Ex. 10. A series of ellipses have their equiconjugate diameters
of the same magnitude. One of these diameters is fixed and com-
mon, while the other varies. The tangents drawn from any point
on the fixed diameter produced will touch the ellipses in points
situated on a circle. (Apply Prop. XXII.)
Ex. 11, If CX, CP ai'e the abscissa and ordinate of a point P on
a circle whose centre is C, and A''^ be taken equal to AP, and be
inclined to it at a constant angle, the locus of Q is an ellipse.
Def. Chords which join any point on an ellipse to the
extremities of a diameter are called supplemental chords.
Proposition XXXI.
Supplemental chords of an ellipse are parallel to conju-
gate diameters.
Join any point Q on the ellipse to the extremities of a
diameter LCM. Then QL and QM are supplemental
chords.
Draw CP, CD parallel to QL, QM respectively; then
they shall be conjugate diameters.
98 GEOMETRY OF CONICS.
Because LM is bisected in Cand CP is parallel to LQ,
OF bisects MQ, [Euc. VI. 2.
and, therefore, all chords parallel to CD. [Prop. X.
Therefore CD bisects all chords parallel to CP, [Prop. XXX.
and is therefore conjugate to CP.
Ex. 1. Prove that for any assumed pair of conjugate diameters
there can be drawn a pair of supplemental chords parallel to them.
Ex. 2. The diagonals of any parallelogram circumscribed to an
ellipse are conjugate diameters. [The diagonals pass through the
centre of the ellipse. Then see Note on Tangent-Properties,
Ex. 1, 3.]
Ex. 3. The diagonals of the rectangle formed by the tangents at
the extremities of the major and minor axes of an ellipse are equi-
conjxigate diameters.
Ex. 4. The tangent at any point Q on an ellipse meets the equi-
conjugate diameters in T and T'. Prove that the triangles OCT
and QCr are as CP : CT"\ [Apply Prop. XXII.]
* Proposition XXXII.
The square of the ordinate of any iwint on an ellipse
with respect to any diameter varies as the rectangle under
the segments of the diameter Tnade by the ordinate.
{QV^:PV.FV=CD^:CP\)
Let QVQ' be a double ordinate with respect to the i
ELLIPSE. 99
diameter PCF, meeting it in V. Let CD be the semi-
diameter parallel to Q V.
Now GP bisects QQ' and therefore all chords parallel to
QV or CD. [Def. and Prop. X.
Therefore CD is conjugate to CP. [Def.
But QV.qV:PV.P'V=CD'':CF\ [Prop. XIL
and QV=qV.
Therefore QV^ .PV. rV= CD^ : CP\
Ex. If QP, QP' meet CD, CP in J/, S respectively, prove that
C2f.CS=CD^
Proposition XXXIII.
IfCP, CD he tvjo conjugate semi-diameters of an ellipse
and ordinates PX, DP be draivn to tlte inajor axis, then
(i) PN:CR = DR'.GN=CB.CA.
(ii) GN-^+CR'' = CA'-.
Let NP and RD produced meet the auxiliary circle in
p and d. Join Cp, Cd, and let the tangents at P and p
meet the major axis produced in T. [Prop. XXV.
Then, because PT is parallel to CD, [Props. X and XIV.
the triangles NPT and RDG are similar.
100 GEOMETEY OF CONICS.
Therefore NT : RC=PN : DR ; [Euc. VL 4.
but PN :DE=pIi: dR. [Prop. XI.
therefore NT :RG=pN: dR,
and the angles pNT and dRC are equal, being right
angles. Therefore, the triangles iVpT and RdC are
similar. [Euc. VI. 6.
Therefore the angles pTiV and dCR are equal.
Therefore pT is parallel to dG and the angle dCP = ihe
angle CpT=a. right angle.
Therefore the angle ^:)C'iV= the angle CdR, each being
the complement of the angle dCR.
Therefore the two triangles p(7iV^ and dCR are equal in
[Euc. I. 2G.
[Prop. XL
Ex, 1. li CQ be perpendicular to PT, prove that
CQ.QT-.CT^^ CiV. FN : CIP.
Ex. 2. If the normal at P meets the major and minor axes in O
and g respectively, prove that
(i) PO:CD=CD:CA,
(ii) Pg.CD=CA.CB,
(iii) PQ.Pg = CD-.
Ex. 3. Prove that if two conjugate diameters be at right angles
to each other, they must be the major and minor axes of the ellipse.
Ex. 4. Prove that
{SP -CAf + {SD - CAf = <75^.
Ex. 5. If the tangent at the vertex A cut any two conjugate
diameters in T'and t, show that AT. At = CB^.
Ex. 6. Apply Prop. XXII. to prove this proposition.
every respect.
Therefore
CR=pN
and
PN:CR = PN:pN
= CB:CA.
Similarly
DR:CN=CB:CA.
Again,
CN^+CR^ = GN^+pN^
= Cp^ = CA\
ELLIPSE. 101
If the taugents at P and D meet the major axis in Taxxd t, it may
easily be shown from the relation
CIt:CX=CT:Ct,
that CN^=CR.Et=AR.A'IL
Then apply Prop. IX.
Proposition XXXIV.
The sum of the squares of any two conjugate semi-
diameUrs is constant. (CT- + CD-=^CA'^ + C£^.)
Let CF, CD be the conjugate semi-diameters, and let
PN, DR be the ordinates to the major axis.
Then P2^:CR = CB:CA. [Prop.XXXIIL
Therefore PN^' : CR' = CB^ : CA\
Similarly DR'' : CN^=CB^ : CA',
therefore P.y^ + DR" : CX^ + CR^ =CJBr-.CA'-,
but CN'- +CR- = CA^; [Prop. XXXIII.
therefore PX'^ + DR^ = CRT-,
therefore CP^ +CD''=CA'-^ CB-. [Euc. I. 47.
Therefore, in the ellipse, the sum of the squares of any
conjugate semi-diameters is constant, being equal to the
sum of the squares of the semi-axis major and semi-axis
minor.
Ex. 1. Find the greatest value of the sum of a ]Miir of conjugate
diameters. [The diameters must be equiconjugate.]
102
GEOMETRY OF CONICS.
[SP+S'P=2CA. Then
Ex. 2. If PG, DH be the normals at P and D, prove that
PG- + DH^ is constant.
Ex.3. Prove that >SP.S'P=CD\
square and substitute.]
Ex. 4. OP, OQ are tangents to an ellipse, and SQ is produced to
meet the directrices in R, It'. Prove that
PR . PR' : QR . QR'= OP^ : Oq\
[If PM and QN be the ordinates, it can easily be shown that
PR. PR ^ MX. MX' ^ SP.S'P
Qli . QR' XX . XX' SQ . JS'Q'
Then apply Ex. 3 and Note on Tangent-Properties, Ex. I., 1.]
* Proposition XXXV.
The area of the parallelogram formed by the tangents
at the extremities of a i^air of conjugate diameters is con-
stant (CD.FF=GA.CB.)
The tangents at the extremities of two conjugate
diameters PGP' and BCD' will evidently form a
parallelogram, [Prop. XIV.
the area of which is four times that of the parallelogram
CDTP, where T is the intersection of the tangents at P
andD.
ELLIPSE. 103
Let the normal at P meet the major axis in G and
DGU in F. Draw the ordinates PN and DR to the
major axis.
Then, since the angles at N and F are right angles,
the angle GPN =i\ie angle GCF =i\iQ angle DOR.
[Euc. I. 15 and I. 32.
Therefore the two right-angled triangles GPK and DOR
are similar.
Therefore PG:CD = PX : CR
= CB : CA, [Prop. XXXIII.
therefore PG .PF:CD .PF=CK-:CA .CB;
but PG . PF= CBT-, [Prop. XXVIIL
therefore CD . PF= CA . CB.
Again, the area of the parallelogram CDTP
= CD . PF= CA.CB = constant,
which proves the proposition.
Ex. 1. Find the least value of the sum of a pair of conjugate
diameters. [The diameters are the major and the minor axis. Cf.
Prop. XXXIV., Ex. I.]
Ex. 2. Prove that the parallelogram formed by the tangents at
the extremities of a pair of conjugate diameters is the least that
can be circumscribed about the ellipse.
Ex. 3. If FG meets the minor axis in cf, prove that
PG.Pg = CI>^.
(Prop. XXVIII. Cf. Prop. XXXIII., Ex. 2.)
Ex. 4 If ST be the perpendicular upon the tangent at P, prove
that SP:ST=CD:CB.
[In the figure of Prop. XXIII.,
SP_S'P _SP+S'P_CA
8Y a'Y' SY+S'T' CK"
where CK is the central perpendicular upon the tangent at P.
Therefore |?=^=SS. 1
SY PF CB A
Ex. 5. Prove that SP. S'P==CIfi. [From Ex. 4
SP.S'P _CI^
SY.S'Y' CB*'
104 GEOMETRY OF CONICS.
Then apply Prop. XXIII. Cf. also Prop. XXXIV., Ex. 3, and
Prop. XXXIII., Ex. 2, along with Prop. XXVI., Ex. 4.]
Ex. 6. If the tangent at P meet the minor axis in T, prove that
the areas of the triangles SPS', STS' are as CD^:ST^. [Cf. Prop.
XXVI., Ex. 1.]
Ex. 7. If DQ be drawn parallel to SP and CQ perpendicular to
DQ, prove that CQ = CB. (See Ex. 4.)
Ex. 8. The tangents drawn from D to the circle on the minor
axis as diameter are parallel to the focal distances of P. (See Ex. 4.)
Ex. 9. If on the normal at P, PQ be taken equal to the semi-
conjugate diameter CD, the locus of (? is a circle whose centre is C
and radius equal to CA - CB. [Apply Prop. XXXIV.]
Miscellaneous Examples on the Ellipse,
1. Find the locus of the point of intersection of any
tangent to an ellipse, with the line drawn from the focus
raakinor a constant angle with the tanj^^ent.
O O C3
[A circle. Cf. Prop. XXIII. Observe that if the vertex of a
triangle of a given species be fixed, while one base angle moves
along a fixed circle, the locus of the other base angle is a circle.]
2. The line drawn parallel to the axis through the
intersection of normals at the extremities of a focal
chord, bisects the chord.
3. 8, S' are the foci of an ellipse ; >S^'jR is drawn equal
to A A' ; the line bisecting R8 at right angles touches the
ellipse. (Newton, Book I., Prop. XVII.)
4. Given a focus, the length of the major axis and two
points on the curve, to construct it. (Apply Prop IV.
Newton, Book I., Prop. XVIII.)
5. Given a focus, the eccentricity, and two tangents, to
construct the curve. (Apply Prop. XXIIL, Ex. 5.
Newton, Book I., Prop. XX.)
6. Given a focus, the eccentricity and two points
ELLIPSE. 105
on the curve, to construct it. (Newton, Book I.,
Prop. XX.)
[The directrix touches the two circles having their centres at
the given points, and radii equal to e times their focal distances,]
7. Given a focus and the eccentricity, to describe an
ellipse touching a given line at a given point. (Newton,
Book I., Prop. XX.)
[Let B be the given focus, and P the given point on the tangent
TPT . (Fig. Prop. XXIII.) Draw >SF at right angles to PT,
and produce it to ^, so that TR= TS. Divide 8R internally and
externally at the points K, L in the ratio BA .AX; the circle on
KL as diameter meets RP in »S".]
8. The rectangle under the perpendiculars let fall from
any point on an ellipse on two opposite sides of an
inscribed quadrilateral is in a constant ratio to the
rectangle under the perpendiculars let fall on the other
two sides.
[The proposition holds if instead of perpendiculars on the sides,
lines are drawn making a constant angle with them. Newton,
Book I., Lemmas XVIL-XTX.]
9. The rectangle under the perpendiculars let fall from
any point on an ellipse on two fixed tangents is in a
constant ratio to the square of the pei'pendicular on their
chord of contact,
10. If two fixed tangents to an ellipse be cut by a
diameter parallel to their chord of contact and by a third
variable tangent, the rectangle under the segments of the
two fixed tangents, intercepted between the diameter and
the variable tangent, is constant.
11. The right line joining the middle points of the
diagonals of a quadrilateral circumscribing an ellipse will
pass through the centre. (Apply Ex. 10 and Prop. XXL,
Ex. 5.)
106 GEOMETRY OF CONICS.
12. If a quadrilateral be circumscribed to an ellipse
the diagonals will intersect on the chord of contact of the
sides.
13. Given two conjugate diameters in magnitude and
position to construct the ellipse.
[Through the extremities P, P, D, D' of the given conjugate
diameters PCP', BCD', draw lines parallel to them, forming the
parallelogram EFOH. Divide the half side DE into any number
of equal parts at E, R\ etc. Divide DC into the same number of
equal parts at /, /', etc. The intersection of PR' and Pr' deter-
mines a point on the ellipse.]
14. Given two conjugate semi-diameters in magnitude
and position, determine the axes.
[Let CPy CD be the conjugate semi-diameters. Draw PR per-
pendicular to CD, and on PR take PQ, P(^ on opposite sides or P,
each equal to CD ; then the axes are in direction the bisections of
the angle QC^, while their lengths are the sum and difference of
CQ, CQ'.]
15. Given two conjugate semi-diameters in magnitude
and position, determine the axes.
[Let CP, CD be the conjugate semi-diameters. Draw PR per-
pendicular to CD, and on it take PQ = CD. On CQ as diameter,
describe a circle, and let 0 be its centre. Join OP, cutting the
circle in E and F ; join CE, CF, and take on CE, CF, CA = FP,
CB—EP. Then CA, CB are the semi-axes sought.]
16. Given two conjugate semi-diameters GP, CD, with
centre G and radius GP describe a circle, and let KK' be j
its diameter at right angles to GP ; then will the axes of
the ellipse be equal to KD ± K'D, and parallel to thej
bisectors of the angle KDK\
17. Any diameter of an ellipse varies inversely as the
perpendicular focal chord of its auxiliary circle.
18. If any rectangle circumscribe an ellipse the peri-
meter of the parallelogram formed by joining the points
ELLIPSE. 107
of contact is twice the diameter of the director circle.
(Prop. XXIV.)
19. Given a focus, the length of the major axis, and
that the second focus lies on a fixed straight line, prove
that the ellipse touches two fixed parabolas having the
given focus for focus.
20. Two given ellipses in the same plane have a
common focus, and one revolves about the common focus
while the other remains fixed ; the locus of the point of
intersection of their common tangents is a circle.
[If H be the second focus of the fixed ellipse, K of the revolving
ellipse, and fc^, 6, their serai- minor axes,
where T is the point whose locus is sought.]
21. TQ, Tq are tangents to an ellipse ; CQ, Cq, QjQf,
CT are joined ; QQ' and CT intersect in V. Prove that
the area of the triangle QCQ varies inversely as
\Tv) '^\GV)'
22. SY, S'Y' are perpendiculars on the tangent at P.
Perpendiculars from Y, Y' on the major axis cut the
circles of which SP, S'P are diametei*s in /, J respectively.
Prove that IS, JS', and BG produced meet in the same
point.
23. An ellipse touches two given lines OP, OQ in P
and Q, and has one focus on the line PQ. Find the other
focus and the directrices.
24. S, S' are the foci of an ellipse ; ^F is perpendicular
on the tangent at P. Prove that S'Y bLsects the normal
at P.
25. GP, GD are two conjugate semi-diameters of an
108 GEOMETRY OF CONICS.
ellipse ; Rr is a tangent parallel to PD ; a straight line
GIJ cuts at a given angle PD, Rr in /, J. Prove that
the loci of / and J are similar curves. [It can easily be
shown that GP :CJ^ = 1: 2.]
26. A system of parallelograms is inscribed in an
ellipse whose sides are parallel to the equiconjugate
diameters. Prove that the sum of the squares on the
sides is constant.
27. OP, OQ are tangents to an ellipse ; GU, GV are the
parallel serai-diameters. Prove that
0P.0Q+GU.GV=0S.08'.
28. P, Q are points on two confocal ellipses at which
the line joining the common foci subtends equal angles.
Prove that the tangents at P, Q include an angle equal
to that subtended by PQ at either focus.
29. The foci of a given ellipse A lie on an ellipse B
the extremities of a diameter of A being the foci of B.
Prove that the eccentricity of B varies as the diameter
of A.
30. Normals at the extremities P and D of two con-
jugate semi-diameters meet in ^. Prove that GK is
perpendicular to PD.
31. If GP, GP' be semi-diameters of an ellipse at right
angles to each other, prove that
Qp-2 ' (jp"2
is constant.
32. Having given the auxiliary circle of an ellipse and
a tangent to the ellipse touching the ellipse at a given
point, find the foci.
i
ELLIPSE. 109
33. Find the locus of the centres of circles cutting a
given ellipse orthogonally.
3-i. An ellipse is inscribed in a given triangle. If one
of the foci is known, show how to find the ellipse and its
points of contact with the sides of the triangle.
35. Two fixed points Q, R and a variable point P are
taken on an ellipse ; the locus of the orthocentre of the
triangle PQR is an ellipse.
CHAPTER III.
THE HYPERBOLA.
DESCRIPTION OF THE CURVE.
Proposition I.
Given the focus, directrix, and eccentricity of a liyper-
bola to determine any number of points on it.
^^^^^^^^^^^^^B ^^^^^^^^^^^^H ^^^^^^^^^IH
Let S be the focus, MXM the directrix, and e the
eccentricity. *
Through S draw 8X perpendicular to the directrix.
Divide SX in A so that
8A=e,AX.
110
HYPERBOLA. HI
Also, in SX produced,* take A' so that
SA' = e.A'X.
Then A and A' are points on the hyperbola and are its
vertices.
Take any point NonA'A produced. Through iV draw
PNP' perpendicular to A A'. With centre S and radius
equal to e . XN, describe a circle cutting PXP" in P and
P'. Then P and P' shall be points on the hyperbola.
Draw PM, P'M perpendicular to the directrix.
Then SP = e.XN [Const
= e.PN,
and SF^e.XN
= e.P'M'.
Therefore P and P' are points on the hyperbola.
In like manner, by taking any other point on A' A pro-
duced, a series of points on the curve may be determined
lying on the right hand side of the directrix.
Again, if N be taken on AA' produced, another series
of points on the curve may be determined lying on the
left hand side of the directrix.
Def. The length of the axis intercepted between the
vertices {A, A') of the hyperbola is called the transverse
axis.
Def. The middle point ((7) of the transverse axis is
called the centre of the hyperbola.
Def. A straight line BCR passing through the centre
and perpendicular to the transverse axis, such that
CB'- = CB"2 = CS^-GA^=SA.SA'
is called the conjugate axis.
* Since e is greater than unity, it is clear that A will lie between
S and JT, and A' without SX on the side remote from S.
112 GEOMETEY OF CONICS.
The conjugate axis, unlike the minor axis of the ellipse, does not
meet the curve at all. (See Ex. 3 below.) Its utility in establish-
ing properties of the hyperbola will appear later on.
Ex. 1. The hyperbola is symmetrical with respect to its axis.
Correspondhig to any point N on the line A' A produced, we get
two points P and P' such that the chord PP' is bisected at right
angles by the axis A' A. [Def.
Ex. 2, Any two right lines drawn from any point on the axis to
the curve on opjjosite sides of the axis, and equally inclined to it,
are equal, and conversely.
Ex. 3. Show that the hyperbola lies wholly outside the lines
drawn through A and A' at right angles to the axis.
In order that the circle may intersect the line PNP, the point N
must be so situated that 8N may not be greater than the radius of
the circle SP, that is, e . XN. It may be shown that this is the
case only when N does not lie between A and A'.
Ex. 4. Hence, the hyperbola consists of two distinct branches
lying on opposite sides of the lines drawn through the vertices at
right angles to the axis.
Ex. 5. There is no limit to the distance to which each branch of
the hyperbola may extend on both sides of the axis, so that the
hyperbola consists of two infinite branches.
It is obvious that the point N may be taken anywhere on the
axis beyond A and A'.
Note. — It will be remembered that the parabola consists of one
infinite branch (Chap. I., Prop. I., Ex. 9) and that the ellipse is a
closed oval (Chap. II., Projx I., Ex. 6).
Ex. 6. In any conic, if PR be drawn to the directrix parallel to
a fixed straight line, the ratio SP : PR is constant.
Ex. 7. If an ellipse, a parabola, and a hyperbola have the same
focus and directrix, the parabola will lie entirely outside the ellipse
and inside the hyperbola. (Cf. Chap. I., Prop. I., Ex. 6 and 7.)
Ex. 8, Prove that the locus of a point of triseotion of an arc ot\
a circle described on a given base is a hyperbola.
Ex. 9. If a circle touches the transverse axis at the focus, and]
passes through one end of the conjugate axis, the portion of the]
conjugate axis intercepted = (7-4^^^.
Ex. 10. Prove that the locus of the point of intersection of two'
tangents to a parabola which cut at a constant angle is a hyperbola.
Let OP, OQ be two tangents to a parabola, cutting at a constant
angle a. Draw 01, OTJ perpendicular to the directrix and SP \
then 0/-,St^(Chap. T., Prop. XIII.), and
OS -.01= OS' '.SU,
HYPERBOLA. 113
which is a constant ratio greater than unity since s.OSP=ir-a.
(Chap. I., Prop. XIX.) The locus of 0 is, therefore, a hyperbola
having the same focus and directrix as the parabola.
Ex. 11. P is any point on a given hyperbola (e=2). Z) is taken
on the axis such that SD = SA'. If A'F meets the latus rectum in
K, find the locus of the intersection of BK and SP. [The circle on
A'D as diameter.]
Ex. 12. The angular point A of a triangle ABC is fixed, and the
angle A is given, -while the points B and C move on a fixed right
line. Find the locus of the centre of the circumscribing circle of the
triangle. [A hyperbola of which A is the focus and BC the
directrix.]
Proposition II.
The hyperbola is symmetrical with respect to the con-
jugate axis and has a second focus (S') and directrix.
Let S be the given focus and MX the given directrix.
Talie any point M on the directrix and through the
vertices A and A' draw AH, A'H' at right angles to AA',
meeting the straight line through M and S at H and H'
respectively.
Describe a circle on HW as diameter, and through M
draw PMP' parallel to AA\ to meet the circle in P and
P'. Then P and P' shall be points on the hyperbola.
H
114 GEOMETRY OF CONICS.
For SH:HM=SA:AX
= e,
and SH' : MH' ==SA': XA'
= e,
therefore SH:HM=SH' :MH\
and the angle HPH' is a right angle ; therefore PH
bisects the angle SPM.
Therefore SP : PM=SH: HM
=SA.AX
= e.
Therefore P is a point on the hyperbola.
Similarly it may be shown that P' is a point on the
hyperbola.
Again, the straight line drawn through 0, the centre of
the circle, at right angles to AA\ will bisect both AA'
and PP" at right angles, and will therefore coincide with
the conjugate axis in position.
The hyperbola is therefore symmetrical with respect to
the conjugate axis.
Hence the two branches of the hyperbola, lying on
opposite sides of the conjugate axis, are such that each is
the exact reflexion of the other. Therefore, if A'S' be
measured off = AS and A'X' = AX, and X'M^^ be drawn at
right angles to X'S, the curve could be equally well
described with S' as focus and X'M^ as directrix. The
hyperbola has therefore a second focus ;Si' and a second
directrix X'My
Ex. Every chord drawn through the centre (7 and terminated by
the two branches is bisected at that point. [From the symmetry of
the figure.]
From this proi)erty the point Cis called the centre bt the curve.
HYPERBOLA- 115
PROPERTIES OF CHORDS AND SEGMENTS OF
CHORDS.
Proposition III.
In tlie hyperbola CA=e.CX (i.)
CS=e.CA (ii.)
CS.CX^CA'- (iii.)
\
»
1
*' 1
t i
.V '^A S
\
\
\
\
We have from the definition
HA^e.AX,
BA'^t.A'X = e.AX\
Therefore, by subtraction,
AA' = e{AX'-AX)
= e.XX'.
Therefore CA=e.CX (i.)
By addition US' = g . UX + A'X)
= e.AA\
Therefore CS=e.CA (il)
Therefore CS. CX = CA'- (iii.)
116 GEOMETRY OF CX)NICS.
Ex. 1. Given the transverse and the conjugate axis, find the focus
and the directrix.
Ex.2. Prove that ^' = ! + £?!•
L/A
Ex. 3. If the line through B parallel to the transverse axis meet
the latua rectum in B, then will the triangles SC'D, SXD be similar.
Ex. 4. Prove that
SX^:AX.A'X=CB^:CA^.
Ex. 5. If any line through- the centre meet the perpendicular
through A to the transverse axis in 0 and the dii'ectrix in E, then
AE is parallel to SO.
Ex. 6. In Prop. I., Ex. 8, find the distance between the centres
of the two hyperbolas which are the loci of the points of trisection
of an arc of a circle described on a given base. [One-third of the
given base.]
Proposition IV.
The difference of the focal distances of any point on a
hyperbola is constant and equal to the transverse axis.
Let P be any point on the hyperbola. Join FS,
PS', and through P draw PMM perpendicular to the
directrices.
HYPERBOLA. 117
Then
SP=e.PM,
and
S^P=e.PM.
Therefore
S^P-SP=€{PM-Pm
= e.MM'
= 6. XX'
= AA'.
[Prop. IIL
Ex. 1 . Show how to construct the hj'perbola mechanicallv.
First Method. — Suppose a bar SQ, length r, to revolve round its
extremity *S' which is fixed. Theu if a string of given length I,
attached to the bar at Q and also to a fixed point S^ be always kept
stretched by means of a pencil at P pressed against it (the part QP
of the string being in contact with the rod), the pencil will trace
out a hyperbola with foci at S and .S", and the transverse axis equal
to (/•-/). For
S'P-^PQ = r
and SP + PQ = l,
,S'i* - <SP = r - Z = constant
It should be observed that I must be less than r and greater than
r-ss:
In the same manner, by making the bar revolve round S as
centre, the other branch of the h%-perbola may be described. Tlie
other branch may also be described by taking the string longer than
the rod by the length (r - 1).
Second Method — Suppose two equal thin circular discs A and B
attached to each other, to rotate in the same direction roimd an axis
through their common centre ; and suppose the two ends of a fine
string (which is wrap{)ed roimd the discs and passing through small
rings at C and D in the plane of the discs, is kept stretched by the
point of a pencil at P) to be wound off from the two discs. The
ciirve traced by P will have the property CP — Z>P= constant, and
will, therefore, be a hyperbola.
1 18 GEOMETEY OF CONTOS.
Ex. 2. Given the foci and the transverse axis to determine any
nnniber of points on the curve.
Describe a circle with centre *S^and any radius r ; describe a circle
with centre S' and rddius = r+.4J'. The two circles intersect in
points on the curve.
Ex. 3. Given a focus, a tangent, and a point on an ellipse, prove
that the locus of the other focus is a hyperbola. [The foci will be
the given point and the image of the focus in the tangent. Chap.
II., Prop. XXIIL]
Ex. 4. Given a focus, a tangent, and two points on an ellipse to
construct the curve. (Newton, Book I., Prop. XXI.)
Ex. 5. Given a focus, two tangents, and a point on an ellipse to
construct the curve. (Newton, Book I., Prop. XXI.)
Ex. 6. Given a focus, the eccentricity, a tangent, and a jx)int on
an ellipse to construct the curve. (Apply Chap. II., Prop. XXIII.,
Ex. 5. Newton, Book I., Prop. XX.)
Ex. 7. The diflerence of the focal distances of any point is greater
than, equal to, or less than the transverse axis, according as the
point is within, ujjon, or without the hyperbola, and conversely.
Ex. 8. Tlie locus of the centre of a circle which touches two fixed
circles is an ellipse or a hyperbola. (Cf. Chap. II., Prop. IV., Ex. 4.)
Ex. 9. Given one focus of an ellipse and two points on the curve,
the locus of the other focus is a liy])erbola.
Ex. 10. A parabola passes through two fixed points, and has its
axis parallel to a given line ; prove that the locus of its focus is a
hyperbola.
Ex. 11. Given the base of a triangle and its point of contact with
the inscribed circle, show that the locus of its vertex is a hyperbola.
Ex. 12. Find the locus of the intersection of the tangents fi-ora
two given points A and B to all circles touching AJi at a given
point C.
[An ellipse when C is outside A and B ; a hyperbola when C is
between A and B, except when CA = CB, in which case the locus is
a right line.]
Ex. 1.3. An ellipse and a liyperbola having the same foci inter-
sect in P. If CA, Oa be their semi-axes major respectively and /'iV
the ordinate of P, show that
CA:CS=C.Y:Ca.
Ex. 14. P is any point on an ellipse, of which CA, CB are the
semi-axes ; CD is the semi-diameter conjugate to CP ; Ch is the
semi-conjugate axis of the confocal hvperbola through P. Prove
that Cm^G\fi = Cb\
Let (7a = semi-transverse axis.
Then Clt^ ==CS^- Ca^ = CaS'^ - ^{SP - STf-
= CS'^ - k{(SP+ S'Pf - ASP . S'Pi
= CIfi- CB'. [Chap. 1 1., Prop. XXXV., Ex. 5.
HYPEEBOLA.
119
Ex. 15. JSV, <S"i" ai-e the focal perpendiculars on the tangent at
any point P of an ellipse. Prove that PT. PY' is equal to the
square on the semi-conjugate axis of the confocal hyperbola
through P. r^ = '^^= ^^
lpt Pi" ;
SY sr
^'PY. PY'
CB
PY PY' s^PYTpT
Apply Ex. 14. Cf. Prop. XXI., Ex. 8.]
Ex. 16. Two adjacent sides of a quadrilateral are given in
magnitude and position ; if a circle can be inscribed on the quadri-
lateral, the locus of the intersection of the other two sides is a
hyperbola.
Ex. 17. Prove that the circle in Prop. I., Ex. 12, always touches
a fixed circle. [Centre is second focus of the hyperbola, radius = trans-
verse axis.]
* Proposition V.
The lotus rectum of a hyperhola is a third lyropor-
tional to the transverse and conjugate axes. (SL = yy-^'j
Let LSL' be the latus rectum,
dicular to the directrix.
Draw LM perpen-
120
GEOMETEY OF CONICS.
Then CS=e.CA.
[Prop. III.
SL = e.LM
[Def.
= e.SX.
Therefore SL.CA^GS. SX
^G8{C8-CX)
= CS^-CS.GX
= CS-'-CA^
[Prop. Ill
= CB'.
[Def.
Ex. Prove this proposition by means of Prop.
III.
Ex.4.
♦Proposition VI.
Any focal chord of a hyperbola is divided hai^moni-
cally by the focus and directrix; and focal chords are
to one another as the rectangles contained by their
segments.
Produce the focal chord PSp to meet the directrix in
D, and dra\v PM and pm perpendicular to the directrix.
'J^hen PD'.pD^ PM : pm ;
but PS=e.PAI,
and pS = e . pwi ; [Drf.
HYPEEBOLA.
121
therefore PD:pD = PS: pS.
Hence Pp is divided harmonically in S and D.
Again, PD, SD, and pD being in harmonic progression,
,PM, SX, and pm are also in harmonic progression. But
! SP : PM=SL : SX = Sp :p7rv=e;
therefore SP, SL, and Sp are also in harmonic progres-
sion. Therefore
^j_2SP.Sp_2SP.Sp
SP+Sp Pp '
therefore the focal chord Pp varies as SP . Sp.
Propositiox YII.
If any chord QQ of a hyperbola intersects the directrix
in D, SD bisects the angle between SQ and SQ'.
First, let Q and Q be on the same branch of the
hyperbola
Draw QM, Q'M' perpendicular to the directrix.
Then, by similar triangles,
= SQ:S(/,
122 GEOMETRY OF CONICS.
Therefore SD bisects the exteriw' angle Q'Sq. [Euc. VI. A.
Secondly, let Q, Q' be on opposite branches of the
hyperbola ; then it may be similarly shown that SD
bisects the interior angle QSQ'. [Enc. VI. 3.
Ex. 1. Prove that a straight line can cut a hyperbola iu two
points only. (Cf. Cliap. I., Prop. X., Ex. 8 ; Chap. II., Prop.
VIII., Ex. 9.)
Ex. 2. If two points Q, Q' on a hyperbola be joined with a
third variable point 0 on the curve, the segment qq' intercepted
on either directrix by the chords QO and Q'O produced, subtends
a constant angle at the corresponding focus.
Ex. 3. Given the focus and three points on a hyperbola, find
the directrix and the axis.
Proposition VIII.
The square of the ordinate of any point on a hyjyerbola
varies as tJie rectangle under the segments of the axis
produced, made by the ordiruite.
{PN^ : AN. A'N = CB' : GA\)
Let PK be the ordinate of any point P on the hyper-
bola. Let PA, PA', produced if necessary, meet the
HYPERBOLA. 123
directrix in D aud If. Join SP, SD, SIX, and produce
PS to meet the curve in p.
Then, from the similar triangles PAN aud DAX,
PX:AX=LX:AX.
Also from the similar triangles PA'N and UA'X,
PX.A'X=UX:A'X;
therefore PN'- : ^.V . A'N= DX . UX .AX. A'X.
Again, SD and SU bisect the angles pSX and PSX
respectively ; [Prop. "VTL
therefore the angle DSU is a right angle, aud
DX . D'X = SX\ [Euc. VI. 8.
Therefore PN'- .AX. A'X= SX'- .AX. A'X.
But the ratio SX-: AX .A'X is constant; therefore the
i-atio PX'^ : AN. A'X has the same .value for all positions
of P.
To determine this constant value we have
SA.AX=CS.CA'y [Prop. III.
therefore SX:AX=CS+CA:GA.
Similarly SX:A'X^CS-CA : CA ;
124 GEOMETRY OF CONICS.
therefore SX^:AX.A'X = CS^-CA^:CA^
= GB^:CA'-, [Def.
therefore PN^ ,AN. A'N= Cm : GA\
Ex. 1. Prove that
PN^ : cm - CA^ = CIP : CA^.
Ex. 2. Having shown that
PJV^' : A^. A'N=SX^ : AX. A'X,
apply Prop. V. to complete the proof. [Make P coincide with the
extremity L of the latus rectum.]
Ex. 3. Prove that
CA-^ CI? ~
Ex. 4. NQ parallel to AB meets the conjugate axis in Q. Show
thai, QB.QE = PN\
Ex. 5. (2 is a point on the curve; AQ, A'Q meet PN in D and
E; prove tliat DN.EN=PN^
Ex. 6. If a jpoint P moves such that PN^ : AN . A'N in a con-
stant ratio, PN being the distance of P from the line joining two
fixed points A, A', and N falling outside AA' ; the locus of P is a
hyperbola of which A A' is an axis.
Ex. 7. PNP' is a double ordinate of an ellipse ; show that the
locus of intersection of AP' and A'P is a hyperbola.
Ex. 8. A circle is described through A, A' and P. If NP
meets the circle again in Q, the locus of (^ is a hyperbola.
Ex. 9. NQ is a tangent to the circle on AA' as diameter ; FM
is drawn parallel to C(^, meeting ^^' in M\ show that MN=CB.
[The triangles PMN, QCN are similar.]
Ex. 10. A chord A P is divided in Q, so that AQ:QP=CA^: CB\
Prove that the line through Q at right angles to QN is parallel
to A'P.
Proposition IX.
The locxis of tfce middle points of any system of
parallel chards of a hyperbola is a straight line passing
through the centre.
Let QQf be one of a system of parallel chords, and V its
middle point.
HYPERBOLA.
125
Draw QM, QM' perpendicular to the directrix ; draw
ST perpendicular to QQ' and produce YS to meet the
directrix in K. Produce QQ' to meet the directrix in R,
'a,ndJQ'mSQ,SQf.
Then SQ:QM=SQ::Q'M'
therefore SQ:Sq = QM :Q'M'
= QR:Q'R;
therefore
SQ' - SQ'- : QR' - Q'R^ = SQ'' : QRr.
But SQ^-SQ'' = QY'--Q'7^' [Euc. l. 47.
= {QT+qY){QY-qY)
= ^.Qq.YV.
Similarly QR^-qR^ = 2.QQ; .RV;
therefore YV: RV=^SQ'' : QR^.
Now, the ratio SQ : QM is constant ; also, the ratio
QM : QR is constant, since QQ' is drawn in a fixed
direction. Therefore SQ : QR is a constant ratio ; there-
fore also YV : RV is a constant ratio for all chords of the
126 GEOMETRY OF CONICS.
system. But as R always lies on a fixed straight line
(the directrix), and Y on another fixed straight line (the
focal perpendicular on the parallel chords), intersecting
the former in K, Vmust also lie on a third fixed straight
line, passing through the same point K.
Again, corresponding to a system of parallel chords in
one branch of the hyperbola, there is in the other branch
another system exactly similar thereto ; and the middle
points of the chords of both the systems must lie on VK,
which therefore divides the two branches symmetrically,
Hence, from the symmetry of the curve about the major
and minor axes, and therefore about the centre, VK must
pass through G.
Hence the diameter bisecting any system of parallel
chords of a hyperbola is a straight line passing through
the centre.
Ex. The diameter bisecting any system of parallel chords meets
the directrix on the focal perpendicular to the chords.
* Proposition X.
If any two parallel chords of a hyperbola he drawn
through two fixed points, the ratio oftlie rectangles under
their segments will he constant, whatever he the directions
of the chords.
Let OPQ be a chord drawn through one of the fixed
points 0, outside the curve.
Produce QPO to meet the directrix in R, and join SR,
SP, 8Q. Draw OU, OV parallel to 8P, SQ respectively ;
and draw OD, PM perpendicular to the directrix.
Then RO:RP=OU:PS
= OD:PM,
HYPERBOLA,
127
but
therefore
Similarly
PS^e.PM;
OU=e.OD.
OV=e OD.
Describe a circle with centre 0 and radius equal to
e.OD, passing through U and F; and draw RT, St
tangents to this circle.
Now, by similar triangles,
OP:OR = SU:RU,
and OQ:OR = SV:RV;
therefore OP .OQ:OR^=SU.SV:RU.RV
= St^:Rr-. [Euc. III.36.
Therefore OP . OQ : St^ = OR'- : RT\
Now for a given direction of the chord OPQ the ratio
OR : OD is constant, and, therefore, also the ratio OR : OT,
since OT=e.OD. Therefore, also, the ratio OR.RT is
constant.
If, now, through another fixed point (/ a parallel chord
128
GEOMETEY OF CONICS.
O'P'Q' be drawn, and similar constructions be made, we
shall have OP.OQ: St^ = O'P' . O'Q' : St'^ ;
therefore OP.OQ: O'F . O'Q' = St^ : St'K
But since the points 0 and 0' arc fixed, the two circles
are fixed in magnitude and position, and, therefore, St and
8t' are constants.
Therefore the ratio OP.OQ : O'P'. O'Q' is constant.
Ex. 1. If a system of chords of a hyperbola be drawn through
a fixed point, the rectangles contained by their segments are as
the parallel focal chords, and also as the squares of the parallel
semi-diameters where they exist. (Apply Prop. VI.)
Ex. 2. The ordinates to any diameter at equal distances from
the centre are equal.
* Proposition XI.
If a circle intersect a hyperbola in four i^oints, their
comTnon chords will he equally inclined, two and two, to
the axis.
Let Q, Of q, gf be the four points of intersection.
HYPERBOLA. 129
Then QO.OQ = qO. Oq'. [Euc. III. 35.
Therefore the rectangles under the segments of the focal
chords parallel to QQ' and qq' respectively are equal,
[Prop. X.
and therefore the focal chords themselves are equal.
[Prop. VI.
They are, therefore, equally inclined to the axis, from
the symmetry of the figure. (See also Prop. I., Ex. 2.)
Therefore, the chords QQ', qq' are equally inclined to
the axis.
In like manner it may be shown that the chords Qq
and Q'q', as well as the chords Qq' and qQ', are equally
inclined to the axis.
PROPEETIES OF TANGENTS.
The student should work out the following exercises as
illustrating the method of deducing tangent properties
from the corresponding chord-properties.
I. Deduce from Prop. X., Ex. 1 : —
1. The tangents to a hyperbola from an external point are pro-
portional to the parallel semi-diameters where they exist, and are in
the subduplicate ratio of the parallel focal chords.
2. If two parallel tangents OP, (/P' be met by a third tangent
at Q, in 0 and (7, prove that
OP:aP'=OQ:(yQ.
II. Deduce from Prop. XI. : —
1. PQ and P(^ are chords of a hyperbola equally inclined to the
axis ; prove that the circle PQQf touches the hyperbola at P.
2. If a circle touch a hyperbola at the points P and Q, show that
PQ is parallel to one of the axes.
III. Deduce from Prop. VII., Ex. 1 :—
1. A tangent to one branch of a hyperbola cannot meet the other
branch.
See also Prop. XII. and XIII.
I
130 GEOMETRY OF CONICS.
Proposition XII.
The tangent to a hyperbola at either end of a diameter
is parallel to the system of chords bisected by the diameter.
Let P'CPV be the diameter bisecting a system of
chords parallel to QQ. Let QQ be made to move
parallel to itself, so that Q may coincide with V. Since
^Fis always equal to QfV, [Prop. X
it is clear that Q will also coincide with V, and the
chord in this its limiting position will be the tangent to
the hyperbola at P.
Ex. 1. The tangent at the vertex is at right angles to the trans-
verse axis.
Ex. 2. The line joining the points of contact of two parallel
tangents is a diameter.
Proposition XIII.
The ^portion of the tangent to a hyperbola at any point,
intercepted between that point and the directrix^ subtends
a right angle at the focus, and conversely.
Also, tangents at the ends of a focal chord intersect on
the directrix.
1
H\T»EEBOLA.
131
First, let any chord QQ' of the hyperbola intersect the
directrix in Z; then SZ bisects the exterior angle QSq.
[Prop. VII.
Now, let the chord QQ' be made to turn about Q until the
point Of moves up to and coincides with Q, so that the
chord becomes the tangent to the hyperbola at Q. In
this limiting position of the chord QQ', since Q and Q"
coincide, the angle QSQ vanishes ; therefore the angle
QSq becomes equal to two right angles. But since SZ
always bisects the angle Q^Sq, in this case the angle QSZ
is a rijrht ancrle.
132 GEOMETRY OF CONICS.
Conversely, let QZ subtend a right angle at S, then it
shall be the tangent to the hyperbola at Q. For if not
and if possible let QZ' be the tangent at Q. Then the
angle Q8Z' is a right angle, which is impossible ; therefore
QZ is the tangent at Q.
Secondly, let Q8q be a focal chord and QZ the tangent
at Q.
Join ZS, Zq.
Then the angle QSZ being a right angle, the angle Z8q
is also a right angle. Therefore qZ is the tangent to the
hyperbola at q. Therefore the tangents at Q, q intersect
on the directrix.
Ex. 1. Tangents at the extremities of the latus rectiini intersect
in X.
Ex. 2. To draw the tangent at a given point of a hyperbola.
Ex. 3. If QZ, qZ meet the latus rectum produced in D and d,
then SB = Sd. (Cf. Chap. II., Prop. XV., Ex. 6.)
Proposition XIY.
If from a point 0 on the tangent at any point P of a
hyperbola perpendiculars OU, 01 he drawn to 8P and
the directrix respectively, then
SI/=e.OI,
and conversely.
Join SZ and draw PM perpendicular to the directrix.
Because ZSP is a right angle, [Prop. XIII.
^*Sf is parallel to OU.
Therefore, by similar triangles,
SU:SP = ZO-ZP
= OI:PM;
but SP = e.PM.
Therefore SU=e.OI.
HYPERBOLA.
Again, for the converse proposition, we have
SU^e.OI,
and SP = e.PM
133
Therefore SU:SP=01 :PM
:=ZO:ZP.
Therefore 0 tT" is parallel to ZS, [Euc. VI. 2.
and the angle PSZ is a right angle.
Therefore PZ is a tangent at P. [Prop- XIIL
Ex. If a perpendiinilar through 0 on the transverse axis meet
the curve in Q and ^, then SU=SQ, and 0lP = 0q.0q.
Proposition XV.
The tangent at any point of a hyperbola makes equal
-angles with the focal distances of the point.
Let the tangent at P meet the directrices in Z
and Z.
Draw PMM' perpendicular to the directrices, meeting
them in M and J/' respectively. Join SP, SZ, S'P, B'Z.
134 GEOMETRY OF CONICS.
Then, in the two triangles PSZ and PS'Z', the angles
PSZ and P8'Z' are equal, being right angles, [Prop. XIII.
and SP:S'P = PM:PM'
= PZ:PZ\
and the angles PZS and PZ'&' are both acute angles.
Therefore the triangles are similar. [Euc. VI. 7.
Therefore the angle SPZ=ihQ angle S'PZ'.
Ex. 1, The tangent at the vertex is perpendicular to the axis.
Ex. 2. Given a focus, a tangent and its point of contact, find tlie
locus of the other focus.
Ex. 3. If FCp be a diameter, and if Sp meet the tangent at /' in
J, tiP=ST.
Ex. 4. If an ellipse and a hyperbola have the same foci, they
intersect at right angles. (See Cha]). I., Prop. XIV., Ex. 4.)
Such Conies are called Con focal Conies.
Ex. 5. If the tangent at F meet the axes in T, t, the angles PSt^
STF are supplementary. [The circle round *S'P*S" obviously passes
through t.']
Ex. 6. If the diameter parallel to the tangent at 1' meet 8P and
S'P in E and E\ the circles about the triangles SCE, S'CE' are
equaL
HYPERBOLA. 135
Ex. 7. Tangents at the extremities of a focal chord FSQ meet in
T. Prove that
2-JPTQ--PS'Q = -2 right angles.
Ex. 8. Y, Y' are the feet of the focal perpendiculars on the
tangent at P ; if PX he the ordinate, the angles PNY, PS Y' are
supplementary. [iJ'NY^ 1.PS Y= WS' r=ir- lPN Y'. ]
Ex. 9. A parabola and a li^-jierbola have a common focus S, and
their axes in the same direction. A line SPQ cuts the curves in P
and Q. If the tangents at P, Q meet in T, prove that lPTQ= \i-SS'Q.
(See Prop. I., Ex. 7.)
Ex. 10. P is a point on a hyperbola whose foci are S., S' ; another
hyperbola is described whose foci are S, P, and whose transverse
axis = 5P— 2P.S'. Prove that the hyperbolas will meet at only one
point, and that they will have the same tangent at that point.
[Apply Prop. IV. If ^ be a point of intersection, QP = QS' + PS' ;
Q, therefore, is the other extremity of the focal chord P.S'.]
Ex. 11. A chord PR VQ meets the directrices in R and F, P, Q
being on ditferent bi-anches. Prove that PR and VQ subtend, each
at the focus nearer to it, angles of which the difference is equal to
the angle between the tangents at P and Q. (Apply Prop. VII.)
Proposition XVI.
To draw tivo tangents to a hyperbola from an external
point.
Let 0 be the external point.
Draw 0/ perpendicular to the directrix, and with
136 GEOMETRY OF CONICS.
centre S and radius equal to e . 01 describe a circle.
Draw OU, OU' tangents to this circle and let US and
SU' produced meet the hyperbola in Q, Q'. Then OQ,
OQ' shall be the tangents required.
For 0C7 is at right angles to 8U, [Euc. III. 18.
and 8U=e.0I
Therefore OQ is the tangent to the hyperbola at Q.
[Frop. XIV.
Similarly OQ' is the tangent at Q'.
Note. — If it had been necessary to produce both SU and SU' in
the same direction, to meet the curve, the points of contact would
have been on the same branch, instead of being on opposite
branches, as in the figure.
Ex. 1. Alternative Construction. — With centre 0 and radius OS,
describe a circle. AVith centre S' and radius equal to the trans-
verse axis, describe another circle intersecting the former in M
and M'. Join S'M and S'M', and produce them to meet the curve
in Q and Q'. OQ, OQ' are the tangents required. (Cf. Chap. II.,
Prop. XVIII., Ex. 1.)
Ex. 2. Prove that only two tangents can be drawn to a hyper-
bola from an external point.
Pkoposition XVII.
The two tangents that can be drawn to a hyperbola
from an external point subtend equal or supplementary
angles at the focus according as the points of contact are
on the same or opposite branches of the curve.
First, let OQ, OQ' be the two tangents from 0, Q and
Q' being on the same branch of the curve.
Join SO, SQ, SQ', and draw 01, 011,011' perpendiculars
upon the directrix, SQ, SQ' respectively.
Then SU = e.01 = SU'. [Prop. XIV.
Therefore OU=OU'. [Euc. 1. 47.
Therefore the angles OSU and OSU' are equal, [Euc. I. 8.
HYPERBOLA.
137
and they are the angles which the tangents subtend at
the focus.
Secondly, let Q and Q' be on opposite branches of the
Then it may be similarly proved that the angles
curve.
OS U and OSU' are equal; therefore the angles OSQ and
OSQ' are supplementary,
Ex. 1. In Fig. 1 prove that 0$, OQ' subtend equal angles at 5*.
Ex. 2. The portion of any tangent intercepted between the
tangents at the vertices, subtends a right angle at either focus.
Ex. 3. Find the locus of the centre of the inscribed circle of the
triangle SQ>S'. [The tangent at the vertex A.]
188
GEOMETRY OF CONICS.
Ex. 4. Show that the chord of contact QQ' is divided harmoni-
cally by SO and the directrix.
Ex. 5. If PN be the ordinate of P, and PT the tangent, prove
that SP:ST=AN:AT.
Ex. 6. Two points P and Q are taken on the same branch of the
curve and on the same side of the axis ; prove that a circle can be
drawn touching the four focal distances. [The centre is the point
of intersection of the tangents at P and Q. Apply Prop. XV. j
* Proposition XVIII.
The two tangents that can he drawn to a hyperbola
from an external point make equal or supplementary
angles with the focal distances of the point according as
the points of contact are on the opposite or same branches
of the curve.
First, let OQ, OQ be the two tangents from 0, Q ana Q'
being on opposite branches of the curve.
Join SQ, SQ\ SO, S'Q', S'Q. S'O, and produce QS to B.
Let H be the point of intersection of SQ' and S'Q.
Then the angle SOQ
= the angle OSR — the angle OQS [Euc. I. 32.
= half the angle Q'SR -half the angle SQS'
[Props. XVII. and XV.
= half the angle SHQ.
HYPERBOLA.
139
Similarly the angle S'OQ'
= half the angle S'FQ'.
Therefore the angle SOQ = the angle S'OQ\
[Euc. 1. 15.
Secondly, let Q, Q' be on the same branch.
Then the angle SOQ
= two right angles —the angle O^S^Q— the angle OQS
[Euc. I. 32.
= two right angles — half the angle Q^S'Q' — half the
angle SQS' [Prop. XVIL and XV.
= two right angles— half the angle SHS'. [Euc. I. 32.
Again, the angle S'OQf
= two right angles — the angle OQ^S' — the angle OS'Q'
[Euc. I. 32.
= half the angle ,S'Q'6"- half the angle QS'Q.
[Props. XV. and XVII.
= half the angle SHS'
Therefore, the angles SOQ and S'OQ' are together equal
to two right angles.
Ex. 1. Tangents are drawn from any point on a circle through
the foci. Prove that the lines bisecting the angle between the
tangents, or between one tangent and the other produced, all
pass through a fixetl point. [A point of intersection of the circle
with the conjugate axis.]
140 GEOMETEY OF CONICS.
Ex. 2. A hyperbola is described, touching the four sides (pro-
duced, if necessary) of a quadrilateral ABCD which is inscribed in
a circle. If one focus lies on the circle, the other also lies on it.
Il.S'CD=lSCB=lSAB=lS'AD.-]
Proposition XIX.
Tlie tangents at the extremities of any chord of a hyper-
bola intersect on the diameter which bisects the chord.
Let QQ' be the chord and qq' any other chord parallel
to it.
Let Qq, Qq' produced meet in 0. Bisect QQ in V and
let 0 V meet qq' in v.
Then QV:qv = OV:Ov
= Q'V:q'v;
but QV=Q'V,
therefore qv = q'v.
Thus OvVia the diameter bisecting the system of chords
parallel to QQ'. [Prop. IX.
If now the chord qq' be made to move parallel to itself
till it coincide with QQ, QqO and Q'q'O will become the
tangents to the curve at Q and Q' respectively. They
thus meet on the diameter bisecting QQ.
HYPERBOLA. 141
Ex. 1. Given a diameter of a hyperbola, to draw the system of
chords bisected br it.
Ex. 2. If a circle passing through anv point P on the curve,
and having its centre on the normal at P, meets the curve again
in Q and R, the tangents at Q and R intersect on a fixed straight
line.
[The tangent at P and QR are equally inclined to the axis (see
Prop. XI.) ; QR is, therefore, fixed in direction.]
Propositiox XX.
If the tangent at any -point Q of a hyperbola meet any
diameter CP in T and if QY he the ordinate to that
diameter, CV.CT=CF'-.
Draw the tangent FB at P, meeting QT in R, and
draw PO parallel to QT, meeting QT" in 0.
Then since POQR is a parallelogram, [Prop. XII.
MO bisects PQ, and therefore passes through the centre G.
[Prop. XIX.
By similar triangles
CV: GP = CO:CR = CP: CT,
therefore CV.CT=CP\
Note. When the diameter coincides with the trans-
verse axis the result is stated thus : —
142 GEOMETRY OF CONICS.
If the tangent at Q meets the transverse axis in T and
QN be the perpendicular' on the transverse axis.
CN.GT=GA^
From this it may be shown that
If the tangent at Q Tneets the conjugate axis, produced
if necessary in t, and Qn he the perpendicular on the
conjugate axis, Cn.Gt= CB^.
QN. Ct _ QN^ QN"^ _ QN"^
GN.CT CN.NT^Cm-CN,CT CN'^-CA^'
. • . Cn.Ct = CIP. [Prop. VIII.
These two results are important, and should be carefully noted
by the student.
Ex. 1. If the tangent at Q meet the transverse axis in T and
QN be the perpendicular on the transverse axis, show that
CN. NT=AN. NA'.
Ex. 2. In Ex. 1, if TD be drawn perpendicular to the axis to
meet the circle described on AA' as diameter, then DN touches
the circle.
Ex. 3. In Ex. 2, prove that
DN:QN=CA:CB.
Also if DA be produced to meet FN in K,
qN:NK=CB:CA.
(Apply Prop. VITI., and see Ex. 1.)
Ex. 4. Any diameter is cut harmonically by a tangent and the
ordinate of the point of contact of the tangent with respect to the
diameter.
Ex. 5. Any tangent is cut harmonically by any two parallel
tangents and the diameter through their points of contact. (Ex. 4.)
Ex. 6. If PN be the ordinate of a point P, and NQ be drawn
parallel to AP to meet CP in Q, AQ shall be parallel to the
tangent at P.
Ex. 7. If the tangent at P intersect the tangents at the ver-
tices and the transverse axis in R, r and T, show that
(i) AT.A'T=CT.TN.
i^ii) AR.A'r^CBK
Ex. 8. P is any point on a hyperbola. Prove that the locus
of the centre {Q) of the circle touching SP, S'P produced, and the
transverse axis, is a hyperbola.
HYPEEBOLA. 143
[Let QM be the ordinate of Q ; then, if the tangents at A and P
meet in F, QSF is a right angle, and
QMSA QM_ SA'
MS~Ar MS AT'
(?JP . Jl^.S' . MS' =SA^: CB». Ex. 7.
Then apply Prop. VIII., Ex. 6.]
Ex. 9. The tangent at P bisects any straight line perpendicular
to AA', and terminated by AP and A' P.
[Let the tangent at P, AP, A'P meet the conjugate axis in t,
E, E respectively. Then
CE- CE CA. A'X- CA' . ^.V_ iCA^ _ 2Ct
PN AN.A'N ~AN.A'N PX'
[Prop. VIII.
.-. CE- CE = 2Ct, or t bisects EE."]
Ex. 10. An ellipse and a hyperbola are describe<l, so that the
foci of each are at the extremities of the transverse axis of the
other ; prove that the tangents at their points of intersection
meet the conjugate axis in points equidistant from the centre.
[The conjugate axes of the two curves are equal in length.}
Proposition XXI.
The locus of the foot of the perpendicular drawn from
either focus upon any tangent to a hyperbola is the
circle described on the transverse axis as diameter ; and
the rectangle under tJie focal perpendiculars on the
tangent is equal to the square of the semi-conjugate aods.
{SY.S'T' = CW)
Let SY, S'Y' be the focal perpendiculars upon the
tangent at any point P.
Join SP, S'P, and produce SY to meet S'P in It
Join CY.
Then in the triangles SPY, RPY, the angle ,SPF = the
angle RPY, [Prop. XV.
the angles SYP, RYP are equal, being right angles, and
YP is common.
144
GEOMETRY OF CONICS.
Therefore SP = PR, SY= YR.
Also SC=CS';
therefore OF is parallel to S'P.
Therefore CY=^S'R
= 1{S'P-PR)
==^{8P-SP)
= CA\
therefore the locus of Y is the circle described on the
traDsverse axis as diameter.
[Euc. I. 26.
[Euc. VI. 2.
[Euc. VI. 4.
[Prop. IV.
Similarly it ma}'- be shown that the locus of Y' is the
same circle.
Again, produce FC to meet S'Y' in y. Then y will be
on the circle.
For, since GS=CS\ and >SfF is parallel to ST, the
triangles SGY, S'Cy are equal. [Euc. I. 26.
Therefore Cy = CY=CA, showing that y is on the circle.
Also SY=8'y,
therefore SY.ST^S'y .S'r = S'A' .S'A [Euc. III. 35.
=SA.SA' = CB''. [Def.
HYPEEBOLA. 145
Ex. 1. If CE drawu parallel to the tangent at P meet S'P in E,
then PE=^C A.
Ex. 2. From a point on the circle on A A' as diameter lines are
drawn touching the curve in P, P'. Prove that SP', S'P are
parallel. [Each is parallel to C'Y.]
Ex. 3. If through any point J' on the circle on AA' as diameter
FP be drawn at right angles to ST, YP will be a tangent to the
hyperbola.
Ex. 4. If the vertex of a right angle moves on a fixed circle, and
one leg passes through a fixed point outside the circle, the other
leg will always touch a h^-perbola.
Ex. 5. Given a focu.s, a tangent, and a ix)int on a hyperbola, find
the locus of the other focus. [An arc of a fixed hyperbola of which
the foci are the given point and the image of the focus in the
tangent.]
Ex. 6. Given a focus, a tangent, and the trausvei-se axis, fijid the
locus of the other focus. [A circle ; centre R, radius = J^ J'.]
Ex. 7. If PX be the ordinate of P, the points T, Y', N, C lie on
a circle.
Ex. 8. The right lines joining each focus to the foot of the per-
pendicular from the other focus on the tangent meet on the normal
and bisect it.
Ex. 9. Alternative Construction for Prop. XYI.
Let 0 be the external point. On OS as diameter describe a circle,
cutting the circle on AA' as diameter in J'and J''. Then Of and
OY' produced will be the tangents required.
Ex. 10. If tangents be drawn from P to a circle described with
<S^' as centre and radius equal to CB, the chord of contact wiU touch
the circle described on AA' as diameter. [The line through y per-
pendicular to S'P will be the chord of contact.]
Ex. 11, If tlie tangent at P cuts the transverse axis in T, prove
th&t AT. A' T=YT. Y'T.
Ex. 12. Find the position of P when the area of the triangle
YCT* is the greatest possible.
[Cr=CF = CJ ; therefore YCY' must be a right angle. If the
tangent at P meets CB in f, PN . Ct=^CBr-. (Prop. XX.) Also the
triangles CYS, CY't are equal ; therefore PS. CS=CB-.]
Ex. 13. If SY, SZ be perpendiculars on two tangents which
meet in 0, OZ is perpendicular to S'O. [S'O is parallel to the
bisector of YCZ. Apply Prop. XVIL]
146
GEOMETRY OF CONICS.
Ex, 14. An ellipse and a hyperbola are confocal ; if a tangent to
the one intersects at ricrht angles a tangent to the other, the locus
of the point of intersection is a circle.
Let SY, S'V be the focal perpendiculars upon the tangent to the
ellipse, and SZ, S'Z' those upon the tangent to the hyperbola ; let
the tangents meet at 0 ; let a, h be the semi-axes of the ellipse, and
a, /3 those of the hyperbola. Tlien if CFbe perpendicular to YOY',
or. or=rr2-or2
and C02 + 0Y.0 Y' = CY"^ = CA^ ;
CO'^ + SZ.S'Z'^a?
or CO'^^a'^-ft\
See also Prop. IV., Ex. 14, 15.
Ex. 15. If an ellipse and a hyperbola are confocal, the difference
of the squares of the central distances of parallel tangents is con-
stant ( = b^ + fr-. Ex. 14.)
* Proposition XXII.
The locus of the intersection of tangents to a hyperbola
which cut at right angles is a circle.
Let the tangents OT, OT' cut at right angles at 0.
Draw SY, GK perpendicular to OT, and 8U, GK' per-
pendicular to OT. Join GY, GU, GO, and produce GK
to meet SU in II.
HYPERBOLA. 147
Now Y and U are on the circle on AA' as diameter ;
[Prop. XXI.
therefore CY=CU=CA.
Now (702 = CK'-+CK'^, [Euc. 1. 47.
and CT'=CK'-+YK^;
therefore CA^= CK'- + SH\
Also Gm=CK'^+UK'\
therefore CA'- = CK' + HC- ;
therefore 2GA'' = CK-^+ CK'^+SH'^+HC^
^CO' + CS-'. [Euc. 1.47.
But GS^=CA^+Cm; [Def.
therefore CO^ = CA^ - CB^.
Hence the locas of 0 is a circle described with centre C.
Xote. — This circle is called the director circle of the hyf)erbola.
In the case when CB is greater than CA, CA'^—CB^ is negative, and,
therefore, the locus does not exist, that is, when CB is greater than
CA the hyjx;rlx)la has no tangents cutting at right angles.
Ex. Four tangents to a hyperbola form a rectangle ; if one side
UV of the rectangle intersect a directiix in /\ and .S" be the corre-
sponding focus, the triangles FSU, FVS are similar.
[SF^ = CF^--CX'' + SX^
= CF-^ + r.S'2 - 2CS. CX= CF^ - CA^+ CB"-
= square of tangent from Fto the director circle
= FU.FV.]
PROPERTIES OF NORiMALS.
Proposition XXI 1 1.
The normal at any point of a hyperbola makes equal
angles with the focal distances of the point
Let the normal UPG at the point P meet the axis
in G.
148 GEOMETEY OF CONICS.
Let PT be the tangent at P. Then
the angle SFT= the angle S'PT [Prop. XV.
But the angles TPG and TPH are equal, being right
angles ; [Def.
therefore the angle ^PG^ = the angle S'PII.
Ex. 1. If the tangent and norn'al at I' meet the conjugate axis
in t and g, P, i, g, S, S' lie on the same circle.
Ex. 2. If a circle through the foci meet two confocal hyperbolas
in F and Q, the angle between the tangents at P and Q is equal to
PSQ.
Ex. 3. The tangent at P meets the conjugate axis in t, and tQ is
perpendicular to SP. Prove that SQ is of constant length.
[If ,SF is perpendicular to Ct, CY= CA. Prop. XXI. Also Q, S,
C, t lie on a circle. .'. LtQC=USS' = UPS'=LtPS. /. CQ \\ SV, and
SQ=Cr=CA.]
Ex. 4. If from cf a perpendicular glC be drawn on SP, show that
PK=CA. (Cf. Cliap. II., Prop. XXVI., Ex. 3.)
Ex. 5. Prove tliat SP . S'P=PG . Pg. (Cf. Cliap. II., Prop.
XXVI., Ex. 4.)
*Pkoposition XXIV.
If the normal at any point P of a hyperbola meet the
transverse axis in G,
SQ = e.SP.
HYPEEBOLA.
U9
Join S'F.
Then, since PG bisects the exterioi* angle between SP
and S'P,
SV : SG = ST : SP ; [Euc VI. A.
therefore S'G-SG:SG = S'P - SP : SP,
or SG : SP = S^G - SG : S'P - SP.
But S'G-SG^SS'=e.AA\ [Prop. III.
and S'P-SP = AA'; [Prop. IT.
therefore SG = e.SP.
Ex. 1. The projection of the normal upon the focal distance of
any point is equal to the senii-latus rectum. (Cf. Chap. II., Prop.
XXVIL, Ex. 4.)
Ex. 2. A circle passing through a focus, and having its centre on
the transverse axis, touches the curve ; prove that the focal distance
of the |)oint of contact is equal to the latus rectum.
Ex. 3. Draw the normal at any point without di-awiug the
tangent.
* Proposition XXV.
The iioi'mal at any 2)oint of a Jiyperhola teiininated
by either axis varies inversely as the central 'perpen-
dicular upon the tangent.
{PG.PF=Cm\ Pg.PF=CA\)
150
GEOMETRY OF CONICS.
Let the normal at P meet the transverse and conjugate
axis in G and g respectively, and let the tangent at P
meet them in T and t respectivel3^
Draw PN, Pn perpendicular to the transverse and
conjugate axis, and let a straight line through the centre,
drawn parallel to the tangent at P, meet NP, GP pro-
duced and Pn in R, F, and r respectivelj'-.
Then, since the angles at N and F are right angles,
G, N, F, li lie on a circle.
Therefore PG . PF= PN . PR [Euc. III. 3r,.
= C)l . Gt [Euc. I. 34.
= CB\ [Prop. XX., Note.
Again, since the angles at n and F are right angles,
g, F, T, n lie on a circle.
Therefore Pg . PF= Pn . Pr [Eua III. 36.
==GN,GT [Euc. I. 34.
= GA^ [Prop. XX., Note.
Therefore both PG and Pg vary inversely as PF, whicli
HYPERBOLA. 151
is equal to the central perpendicular upon the tangent
at P.
Ex. In Prop. XXIII., Ex. 1, prove that
Gg=e.Sg
Apply Prop. III., Ex. 2.
* Proposition XXVI.
// tlie normal at any point P of a hyperbola meet the
transverse axis in G, and PX he the ordinate to that
axis.
(i) GN:GN=CE':CAK
(ii) CG = e^.CN
Let the normal meet the conjugate axis in g. Draw
Pn perpendicular to the conjugate axis, and CF parallel
to the tangent at P.
Then, because the triangles PNG and Png are similar,
GX :CN = PG: Pg [Euc. VI. 2.
= PG.PF.Pg.PF
= CB'-:CA-'; [Prop. XXV.
therefore 6'.V+ GN : CX =CA^-{-CJr-. CA\
or GG:CN = CS^'.CA\ [Def.
152 GEOMETEY OF CONICS.
But GS=e.CA; [Prop. III.
therefore CG = e\CN.
Ex. 1. Prove that
CG . Cn : Cg . CN= CIP : CAl
Ex. 2. Show that
Sn:Cn=CA^:CIP.
Ex. 3. If the tangent and normal at P meet the axis in 7' aiul ''/',
prove that
(i) lYO . CT= CB\ (ii) CG . CT= CSK
[Apply Prop. XX.]
Ex. 4. Find the locus of the points of contact of tangents to a
series of confocal hyperbolas from a fixed point on the axis.
[From Ex. 3 (ii), G the foot of the normal is fixed ; hence P lies
on the circle of which TG is diameter.]
PROPERTIES OF ASYMPTOTES.
Def. When a curve continually approacnes to a fixed
straight line without ever actually meeting it, but so that
its distance from it, measured along any straight line,
becomes ultimately less than any finite length, the fixed
straight line is called an asymptote to the curve.
Proposition XXVII.
The diagonals of the rectangle formed by perpen-
diculars to the axes of a hyperbola, drawn through
their extremities, are asymptotes to the curve.
Let GR, CR' be the diagonals of the rectangle formed
by perpendiculars through the extremities A, A', B, B' of
the axes of the hyperbola. Through any point N on the
transverse axis draw pPNP'p' perpendicular to it, meet-
ing the curve in P and P', and CR, GR' in p, p'
respectively.
Now Pm : AN . A'N= CB^ : GA'', [Prop. VIII.
or PJ\^2 . CM^ _ (-1^2 ^ c^2 . 0^2 j^j,,,^ jj g
HYPERBOLA.
153
Again pm : CN'- = AR^:CA'
therefore pN^ - PN^ : CA^ = CR^ : GA\
or ■pN-^-PN^ = CB~
But since pp' is bisected in N,
pN'- - PN-' =pP . pP. [Euc. II. 5.
Therefore pP . p'P = GB-.
Now pP = ]sP^N^\
and NP- varies as AN . A'N, [Prop. VIII.
and Np' varies as CN.
Hence, as N moves along A A produced, both NP and
N'p, and therefore also Py , continually increase. But
the product pP . p'P, of which one factor p'P continuaDy
increases, is constant ; therefore p'P continually dimin-
ishes, and becomes ultimately less than any finite length,
however small. CR, therefore, is an asymptote to the
hyperbola. Similarly, CR is another asymptote.
Ex. 1. The liues joining the extremities of the axes are bisected
by one asymptote and parallel to the other.
Ex. 2. Any line parallel to an asymptote cannot meet the curve
in more than one ptiint.
Ex. 3. Prove that the angle l>etween the asymptotes of the
154 GEOMETRY OF CONICS.
hyperbola in Prop. I., Ex. 10, is double the exterior angle between
the tangents.
Ex. 4. The circle on A A' as diameter cuts the directrices in tlie
same points as the asymptotes.
Ex. 5. If the directrix meets CM in F, prove that (i) CF=AC ;
(ii) OFS is a right angle.
Ex. 6. Given one asymptote, the direction of the other, and the
position of one focus, find the vertices.
Ex. 7. If CR meets the directrix in F, AFis parallel to SR.
Ex. 8. Given the asymptotes and a focus to find the directrix.
[Apply Ex. 5 (ii).]
Ex. 9. Given the centre, an asymptote, and a directrix, to find
the focus. [Apply Ex. 5 (ii).]
Ex. 10. Given an asymptote, the directrix, and a point on the
hyperbola, to construct the curve. (Ex. 5.)
Ex. 11. The straight line drawn from the focus to the directrix,
parallel to an asymptote, is equal to the semi-latus rectum, and is
bisected by the curve. (Cf. Ex. 13.)
Ex. 12. The perpendicular from the focus on either asymptote is
eqvial to the semi-conjugate axis.
Ex. 13. The focal distance of any point on the curve is equal to
the length of the line drawn from the point parallel to an asymjitote
to meet the directrix. (Cf. Ex. 11.)
Ex. 14. Given the eccentricity of a hyperbola, find the angle (6)
between the asymptotes. ( sec _ = e. j
Ex. 15. Prove that the tangents to a hyperbola from C coincide
with the asymptotes.
Apply Prop. XVI., Ex. 1, obsex'ving that the tangents are unes
bisecting S3f, SM' at right angles.
The asymptotes may thus be regarded as tangents to the hyper-
bola whose points of contact are at infinity.
Ex. 16. If the tangent at P meets an asymptote in T, prove that
ST will bisect the angle between PS and the line through /S parallel
to the asymptote. (Apply Ex. 15 and Pi-op. XVII.)
Ex. 17. If the tangent at P meets an asymptote in 7' prove that
lSTP^i.S'TC=^lPS'T. (Ex. 15.)
Ex. 18. If a tangent meet the asymptotes in L and 3/, the angle
subtended by LM at the farther focus is half the angle between the
asymptotes.
[Apply Ex. 16 and Prop. XVIII. If S'L, S'M' be drawn parallel
to the asymptotes, LS', MS' bisect the angles PS'L', PS'M'.\
HYPEEBOLA. 155
Ex. 19. Given an asymptote, the focus, and a point on the hyper-
liola to construct the curve.
[The feet of the focal i>erpendiculars on the asymptote and the
tanffent at the point (Ex. 16) will lie on the circle described on AA'
as diameter (Ex. 15 and Prop. XXI.), •whence the centre is deter-
mined ; the directiix is found at once by Ex. 5.]
Ex. 20. The tangent and normal at any point meet the asymptotes
and the axes respectively in four points lying on a circle, which
passes through the centre of the hyperbola, and of which the radius
varies inversely as the central perpendicular on the tangent.
Ex. 21. The radius of the circle which touches a hyperbola and
its asymptotes is equal to the part of the latus rectiun intercepted
between the curve and an asymptote. (Apjjly Prop. Y.)
Ex. 22. A pai'alH>la /' and a hyperbola S have a common focus,
and the a.symptotes of H are tangents to P- Prove that the tangent
at the vertex of /* is a directrix of JI, and that the tangent to Pat
it« intersection with X passes through the farther vertex of S.
[The line joining the feet of the focal perpendiculars upon the
as^-mptotes is the tangent at the vertex of P (Chap. I., Prop.
XXIII.), and the dii-ectrix of H (Ex. 5). If P be a common
point, and PM be perpendicular to the directrix of ff, we have
SP:PM=SC:CA, and SP=PAf+SX. .: SP : SA'=CS : AS.
.-. SP. AS=SX. CS= CB-^ = SA . SA'. .-. SP=SA' and A'P touches
the parabola at P. (Chap. I., Prop. XIY.).]
Ex. 23. If an ellipse and a confocal hyperbola intersect in P, an
asymptote passes through the point on the auxiliary circle corre-
sponding to P. (ApplyProp. lY., Ex. 13.)
Proposition XXVIII.
If through any yoint on a hyperbola a straight line
parallel to either axis he draivn meeting the asyraptotes,
the rectangle under its segments is equal to the square of
the semi-a^is to which it is parallel.
First case.
Through any point P on the hyperbola draw Ppp'
parallel to the transverse axis, meeting the asymptotes in
p and J) and the conjugate axis in n.
Then, since pp' is bisected at /*,
Pp . Pp = Pn- -pn?. [Euc IL 6.
156
GEOMETRY OF CONTCS.
Now
PiY^ . ^^_ ^'x= CB'- : CA\
[Prop. VI rr.
therefore
PN^:
:CN-'-GA^=CB^:GA^;
[Euc. II. 6.
or
Cn^
:Pn^-CA'-=CB^:CA\
but
Cn^:pn^ = CB^'.BR^
= CB^:CA\
therefore
PN-^-GA''=pn'',
or
Pn^-pn^ = GA^
therefore
Pp.Plj'=GA\
Second case.
Through P draw qPq parallel to the conjugate axis,
meeting the asymptotes in q, q.
Then, as before,
PN^ : GN^ - GA^ = GB"- : GA\
or PN^ + GR' : Gm = CB"- : GA\
or PN-^ + GB' : Pn^ =GB^:GA^;
but qN^:Pn'- = qN^:GN-'
= AR^:GA''
= GB':GA-^l
therefore q^'' = PN' + GB^,
or qN-^-Pm=GB\
or Pq . Pq = GBT-. [Euc. II. 6.
HYPEEBOLA. 157
I'ROPOSITION XXIX.
If tlirourjh any point on a hypevhola lines he drawn
'parallel to the asymptotes, ilie rectangle under the segments
intercepted between the point and the asymptotes is con-
stant.
Through any point P on the hyperbola draw PH, PK
parallel to the asymptotes, meeting them in H, K. Draw
RAK and 5P5' perpendicular to CA
Then, Ly similar triangles,
PH.Pq = CR':RR,
and PK:Pq=CR:RR,
therefore PH.PK.Pq. Pq = CR' . CR : RR\
or PH.PK.GB^ = CRr-ARA^. [Prop. XXVIII.
= CS--ACE^. [Def.
or PH.PK=\CS^.
Ex. 1. Find the locus of the poiut of intersection of the medians
of the triangle formed by a tangent with the asymptotes. [A
hyperbola having the same asymptotes.]
Ex. 2. P, Q are points on a hyperbola. PL, QM are drawn
parallel to each other to meet one as>Tuptote ; PR, QN are drawn
also parallel to each other to meet the other asjTuptote. Prove
t\\a.t PL . PR=QM .QN.
158 GEOMETRY OF CONICS.
Ex. 3. If through P, P' on a hyperbola lines are drawn ])arallel
to the asymptotes, forming a parallelogram, one of its diagonals
will pass through the centre.
Ex. 4. If P be the middle point of a line which moves so as to
form with two intersecting lines a triangle of constant area, the
locus of P is a hyperbola.
Ex. 5. If through any point of a hyperbola, lines be drawn
parallel to the asymptotes meeting any semi-diameter CQ in P
and R, then CP. CR==CQ\
Ex. 6. A series of hyperbolas having the same asymptotes is
cut by a fixed straight line parallel to one of the asymptotes, and
through the points of intersection lines are drawn parallel to the
other, and equal to either axis of the corresponding hyperbola ;
prove that the locus of their extremities is a parabola.
Ex. 7. Given the asymptotes and a point on the curve, to con-
struct it. (Apply Prop. XXVII., Ex. 5.)
Ex. 8. If a line through the centre meets PH, PK in Z7, T^, and
the parallelogram PUQ V be completed, prove that Q is on the curve.
[If QC, VQ meet the asymptotes in 17\ V, since the parallelo-
grams BK, U' V are equal, PIT . PK= QU'.Q F'.]
Ex. 9. The ordinate JVP at any point of an ellipse is produced
to Q, such that JVQ is equal to the subtangent at P. Prove that
the locus of <2 is a hyperbola.
[If P is on the quadrant AB, the asymptotes are CB and the
bisector of the angle ACB'.]
Ex. 10. If a straight line passing through a fixed point C,
meets two fixed lines OA, OB in A, B, and if P be taken on AB
such that CP^ = CA . CB, find the locus of P.
[Through C draw CB, CE parallel to OA, OB, to meet them.
Through P draw lines parallel to OA, OB meeting CE in K, and
DC in //. Then OD . OE=PE. PK. The locus of P is, therefore,
a hyperbola of which the asymptotes are CH, CK."]
Def. Two hyperbolas are said to be conjugate when
the transverse axis of each coincides with the conjugate
axis of the other.
Thus, a hyperbola which has CB and CA for transverse and
conjugate axes respectively, is called the Covjugate hyperbola^ with
reference to the one we have been dealing with.
The conjugate hyperbola has the same asymptotes as the original
one, since they are the diagonals of the same rectangle. It is
evident that a pair of conjugate hyperbolas lie on opposite sides
of their common asymptotes.
HYPEBBOLA.
159
It has already been pointed out that the tico branches of a
hyperbola togeUier constitute one complete curve ; but it must
not, by analogy, be supposed that a pair of conjugate hyperbolas
together constitutes one entire curve. They are a pair of totally
distinct hyperbolas, although one is of use in deducing some pro-
perties of the other.
Ex. 1. Tangents TP, TQ are drawn to a hyperbola from any
point T on one of the branches of the conjugate. Prove that PQ
touches the other branch of the conjugate.
[(77' bisects PQ in V, Prop. XIX.; and CT.CV=Cr-. Prop.
XX]
Ex. 2. An ordinate NP meets the conjugate hyperbola in Q ;
prove that the normals at P and Q meet on the transverse axis.
[If the normal at Q meets the axes in G and G',
QG'CA^Cy
QG~cm~yG^
Apply Props. XXV., XXVI.]
Proposition XXX
If tlirough any point on a hyperbola or its conjugate
a straight line be drawn in a given direction to meet the
asymptotes, the rectangle under its segments is constant.
Let P be the point on the given hyperbola and Q a
point either on the same hyperbola or its conjugate.
160 GEOMETRY OF CONICS.
Draw pPiTi' and qQq in the given direction, meeting the
asymptotes in p, p' and q, q' respectively. Through P, Q
draw uPu', vQv' parallel to the conjugate axis, meeting
the asymptotes in u, u' and v, v respectively.
Now, by similar triangles,
Pp:qq = Fu-qv,
and Pp':Qq' = Pu':Qv',
therefore Pp . Pp : Qq . Qq' = Pu . Pu' :Qv. Qv;
but Pu . Pu' = GB"^ = Qv . Qu, [Prop. XXVIIT.
therefore Pj) . Pp = Qq . Qq.
Ex. 1. Prove that
Pp.Pp' = Qq.Qq' = CD\
where CD is the parallel semi-diameter terminated by the curve
or its conjugate.
Ex. 2. An ordinate QV of any diameter CP is produced to
meet the asvmptote in R, and the conjugate hyperbola in §'.
Prove that ' QV^ + q'V^ = 2RV\
Prove also that the tangents at Q, Q' meet CP in points equidistant
from C. [Q' V^ -RV^ = C])\ For the second part, apply Prop. XX.]
Proposition XXXI.
If any line cut a hyperbola the segments intercepted
between the curve and its asyTnptotes are equal, and the
portion of any tangent intercepted between the asymptotes
is bisected at the point of contact.
Let any line meet the curve and its asymptotes in Q,
Q' and q, q respectively.
Now Qq . Qq = qq . Qq. [Prop. XXX.
or Qq . QQi + Qq . Q Y = Qcf . QQ' + Qq . Q^q', [Euc. II. 1
or Qq.QQ' = QV'QQ\
therefore Qa = Q'q'.
HYPEEBOL^V. 161
If now QQ' be made to move pamllel to itself until the
points Q, Q' coincide at a point P on the curve it becomes
the tangent to the curve at P and Pp=Pp'
Ex. 1. From a given point on a hyperbola, draw a sti-aiglit line
such that the segment intercepted between the other intersection
-vvith the hyperbola and a given asymptote, shall he equal to a
iriven Hue.
Wlien does the problem become impossible ?
Ex. 2. The foot of the normal at P is eqiudistant from p, p'.
Ex. 3. Prove that Qq . Qq' = Pj^.
Ex. 4. If QK be dl•a^^^l parallel to C'q' and (/K' parallel to Cq,
then Kq = K'(/y ami KQ = K'q'.
Ex. 5. The tangent at P meets an asymptote in 7', and a line
TQ di-awn parallel to the other asymptote meets the curt-e in Q ;
if PQ produced meets the asjTnptotes in jff, R, prove that RR is
trisected at P and Q.
Ex. 6. The diameter bisecting any choitl QQ' of a hyperbola
meets the curs-e in P ; and QH, PK, (/W are drawn parallel to
I (lie a-svmptote meeting the other in H^ A', H'. Prove that
CH.CH=CK\
Ex. 7. A line drawn through one of the vertices of a hyperbola,
and terminated by two lines drawn through the other vertex
I»arallel to the asymptotes, will be bisected at the other point
where it cuts the hyperbola.
Ex. 8. If ^rr be the tangent from q, and QH, TK, qH' l>e
.h-awni parallel to Cq meeting C(i in H, K, H\ prove that
QH+qH' = ^TK.
162
GEOMETRY OF CONICS.
Ex. 9. Through any ])oint P on a hyperbola lines are drawn
parallel to the asymptotes, meeting them in M and N ; and any
ellipse is constructed having CM, CN for semi-diameters. If CP cut
the ellipse in Q, show that the tangent to the ellipse at Q is parallel
to the tangent to the hyperbola at P. [Each is parallel to J/iV.]
* Proposition XXXII.
The area of the triangle forraed hy the asymptotes and
any tangent to a hyperbola is constant.
Let the tangents at the vertex A and at any point P
meet the asymptotes in R, R' and T, t respectively.
Draw PH, PK parallel to the asymptotes, meeting them
in H and K.
Then, since Tt is bisected at P,
CT=2.CH,
and
therefore
CT\
[Prop. XXXI.
[Euc. VI. 2.
Ct = 2,CK,
Ct = 4.CK.CH
= ^.PH.PK
= GR.CR'.
Therefore the triangle CTt is equal to the triangle CRR',
[Euc VI. 15.
8nd is, therefore, constant.
[Prop. XXIX.
[Def.
HYPERBOLA. 103
Ex. 1. If auy two tangents Iw drawn to a hyperbola, the lines
joining the points where they met the asymptotes will be parallel.
Ex. 2. If TOt, T'Ot' be two tangents meeting one asjTnptote in
T. T\ and the other in t, t\ prove tliat
TO.Ot=fO:TO.
Ex. 3. Tangents are drawn to a hyperbola, and the portion of
• ;icli tangent intercepted between the a-symptotes is di^-ided in a
'instant ratio. Prove that the locus of the points of section is a
In-perbola. (Apply Prop. XXIX.)
PROPERTIES OF CONJUGATE DIAMETERS.
Proposition XXXIII.
// one diameter of a hyperbola bisects chords parallel
to a secoTid the second diameter bisects chords pandlel to
the first.
Let GP bisect chords parallel to CD, then CD bisects
chords parallel to CP.
Draw AQ parallel to CD meeting CP produced in 1".
Join A'Q, intersecting CD in U.
Then, because AQ is bisected in Fand AA' \n C, CVh
parallel to A'Q. [Euc. VI. 2.
Again, since A A' is bisected in C and CD is parallel
to AQ, A'Q is bisected by CD. [Euc. YI. 2.
164 GEOMETRY OF COXICS.
Therefore CD bisects all chords parallel to A'Q,
[Prop. IX.
and, therefore, all chords parallel to CP.
Def. Two diameters so related that each bisects chords
parallel to the other are called conjugate diameters.
Thus CP and CI) are conjugate to each other ; so also are the
transverse and the conjugate axes.
It is clear that of two conjugate diameters, one (as CF) will
meet the hyperbola, and the other (as CD) the conjugate
hyperbola.
The portion CD terminated by the conjugate hyperbola is
usually called the semi-diameter conjugate to CP.
Ex. 1. If any tangent to a hyperbola meet any two conjugate
diameters, the rectangle under its segments is equal to the
square of the parallel semi-diameter. (Cf. Clia]>. II., Proj).
XXX., Ex. 7.)
Ex. 2. Given in magnitude and position any two conjugate
semi-diameters of a hvperbola, find the transverse and conjugate
axes. (Cf. Chap. II., Prop. XXX., Ex. 8.)
Ex. 3. Draw a tangent to a hyperbola parallel to a given
straight line.
[The p(iint of contact {P) of the required tangent is obtained by
drawing CD parallel to the given straight line, and CP parallel
to the tangent to the conjiigate hyjaerbola at Z>.]
Ex. 4. If CQ be conjugate to the normal at P, CP is conjugate
to the normal at Q.
Ex. 5. OP, OQ are tangents to a hyperbola from 0. Prove
that CO, PQ are parallel to a pair of conjugate diameters. (Prop.
Ex. 6. An ellipse or a liyperbola is drawn touching the asymp-
totes of a given hyperbola. Prove that two of the chords of
intersection of the curves are parallel to the chord of contact of
the conic with the asymptotes.
[If PP' be the chord of contact and CV bisect PF, then CV,
PP are parallel to a pair of conjugate diameters in both conies.]
Def. Chords which join any point on a hyperbola to
the extremities of a diameter are called siq^plemental
chords.
HYPERBOLA.
165
Proposition XXXIV.
Siipplemental chords of a hyperbola are parallel to
conjugate diameters.
I
Join any point Q on the hyperbola to the extremities
of a diameter LCM. Then QL and QM are supplemental
chords.
Draw CP, CD parallel to QL and QM respectively, then
they shall be conjugate diameters.
Because LM is bisected in C, and CP is parallel to ZQ,
CP produced bisects MQ, [Euc VL 2.
and, therefore, all chords parallel to CD. [Prop. IX.
Therefore CD bisects all chords parallel to CP,
[Propi XXXIIL
and is, therefore, conjugate to it.
Proposition XXXV.
The tangents at the extremities of any pair of con-
jugate diameters meet on tlie asymptotes, and the line
joining the extrenfiities is parallel to one asymptote
and bisected by the other.
IGG GEOMETRY OF CONICS.
Let CP, CD be a pair of conjugate semi-diameters.
]lraw 7'Pr' the tangent at P, meeting the asymptotes in
r and r'. Join Dr and produce rD to meet the other
asymptote in K.
Now, since P is a point on the curve and D on its
conjugate, and DC meets both the asymptotes in C and is
parallel to Pr, [Props. XII. and XXXIII.
DC' = Pr.Pr [Prop. XXX.
= Pr~; [Prop. XXXI.
therefore CD = Pr.
Therefore Dr is parallel to CP, and Cr, PD bisect each
other at 0.
Again, since Pr = Pr\ [Prop. XXXI.
and Or=^OC,
therefore PD is parallel to r'K. [Euc. VI. ±
Therefore Dr = DK, [Enc. VI. 2.
and KDr is the tangent at D. [Prop. XXXI.
Ex. 1. If PD be drawn parallel to an asymptote to meet the
conjugate hyperbola in D, CP, CD are conjugate diametei-s.
Ex. 2. Conjugate diameters of a liyperbola are also conjugate
diameters of the conjugate hyperbola.
HYPERBOLA. 167
Ex. 3. CP, CD are conjugate diameters of a hyperbola. PX,
DJf are ordinates to the transverse axis. Prove that
(i^ CJf.py=CA:CR
{{{) DM:CX^CB:CA.
Let the tangent to the hyjierlxila at P and to the conjugate at
Z>, meet the transverse axis in T, t respectively. Then VP, PT
are parallel to Dt, DC. Now
CT. C'X= CA^=Ct . CM. (Prop. XX.)
CM: CN= CT : Ct = PT:CD= PX : DM= CX : Mt ;
CX*-^CM.Mt = CA^-+CM\ (Prop. XX.)
CM-^ = rX^-CA\
But PT2 : CX- - CA- = CB- : CA^ (Prop. VIIL }
.'. (i) follows immediately.
Ex. 4. If the normal at P meet the axes in G, g, prove that
(i) PG:CD=CB:CA.
(iij Pq:CD=CA:CB.
Cm) PG.P(t=CD^.
[The triangles DCM and PGX are similar, as also the triangles
I )CM And Pgn.'l
Ex. 5. A circle is drawn touching tlie ti*ansverse axis at C,
and also touching the curve. Prove that the diameter conjugate
to the diameter through either p^int of contact, is equal to S^.
[If the normal at P meets the axes in G, g, and the tangent at
P meets CB in t, Ct = PG, and CTf- = PG . Pa = Ct . Cg = CS\ Prop.
XXIIL, Ex. 1.]
Ex. 6. The area of the parallelogram formed by the tangents
at the extremities of any jiair of conjugate diameters, is constant
and equal to A.CA. CB. ' (Apply Prop. XXXIL)
Ex. 7. The tangent at a point P of an ellipse (centre 0) meets
the hyperbola having the same axes as the ellipse, in C and D.
If ^ be the middle point of CD, prove that 0^, OP are equally
inclined to the axes.
[Draw OrR parallel to PQ, meeting the ellipse and hj.-perbola
in r and R ; then OP, Or are conjugate in the ellipse, and OQ, OR
in the hyperbola. If PX, Q3f, H, RL be the ordinates, we have,
for the ellipse.
(Ex. 3.)
PX
OK-
01
Ol.
OX"
~0A-
rV
Similarly
for the hyperbola,
QM
OB*-
01
OM
^OA^'
rV
.•.
PX : 0X=
= QM:OM.]
1()8 GEOMETRY OF CONICS.
Ex. 8. With two conjugate diameters of an ellipse as as}i)ii)-
totes, a pair of conjugate hyperbolas is described. Prove that if
the ellipse touch one hyperbola, it will also touch the other.
[The diameters drawn through the jioints of contact are cf)n-
jugate to each other.]
Ex. 9. Apjily this proposition to prove Prop. X.
1*ROPOSITION XXXVI.
The difference of tlte squares of any tivo conjugate
semi-diameters of a hyperbola is constant
Let CP, CD be a pair of conjugate semi-diameters.
Draw the ordinate qPNq', meeting the asymptotes in
q, (fy and join P7) ; lot PD meet the asymptote in K.
Join Pq.
Then, since the asymptotes are equally inclined to the
ordinate qPNq, [Const.
and PK is parallel to the asymptote Cq', [Prop. XXXV.
the angles KqP and KPq are equal.
Therefore Kq = KP = KD. [Prop. XXXV.
Therefore the circle described on PD as diameter passes
through q, and the angle PqD is a right angle, [Euc. III. 31.
HYPERBOLA. 169
ir, tlieiefore, qD produced meet the conjus^ate axis in M
aud the asymptote Cq in q", qMq" will be at right angles
to CB.
Now Cq- - C'P- = qX- - PN- [Euc. I. 47.
= Pq . Pq [Euc. IL 5.
= CB^, [Prop, xxvin.
and Cq'- - CD' - qM^ - mf- [Euc. I. 47.
= Dq.Dq" [Euc. II. 5.
= CA- ; [Prop. XX\T[II.
therefore GP"- - CD' = CA^ ~ CR-.
Ex. 1. If from any point on an asymptote of a hyperbola,
•ndinates he di-awn to the cui-ve and its conjugate, meeting them
in P and D resi>ectively, show that CP and CD ■will be conjugate
.semi-diametei-s, and conversely.
Ex. 2. Apply Pi-op. XXXY., Ex. 3, to prove this i>roposition.
We have CS- - CJJ ^ = CA-.
Similarlv, if J'li, Dm be ordinates to CB,
Cm--Cn- = CI?,
or DJP-PiY^- = CJP. .
Subtmcting, CP^ ^CD^ = CA- « CBK
Ex. 3. The difference Ijetween the sum of the squai'es of the
distances of any point on the curve from the ends of any diameter,
and the sum of the squares of its distances from the ends of the
conjugate, is constant. [ = 2(CA^ « CZT-).]
Ex. 4. <r is the focus of the conjugate hv)>erbola Iving on CB.
Prove that <rD- SP=CA-CB.
(Apply Ex. 1, and Prop. XXVII., Ex. 5 and 13.)
Ex, 5. Prove that SP.SP^CD^.
[SP''S'P=2. CA. Then square and substitute. Cf. also Prop.
XXIIL, Ex. 5, and Prop. XXXY., Ex. 3.]
Ex. 6. In Prop. XXIIL, Ex. 1, prove that
St'.tg=CB:CD,
CD being conjugate to CP. [Apply Ex. 5 and Prop. XXL]
Ex. 7. If the tangent at P meet any conjugate diameters in
T and t, the triangles ,SPT, S'Pt are similar.
[SP : PT^Pt : S'P. Apply Ex. 5 and Prop. XXXIIL, Ex. L]
Ex. 8. If the tangent at P meet the conjugate axis in t, the
areas of the triangles SP/^, SIS' are the ratio of CD- -.St^. (Apply
Prop. XXIII.. Ex. 1.)
170 GEOMETEY OF CONICS.
Ex. 9. Through C a line is drawn parallel to either focal dis-
tance of P ; if DE is drawn perpendicnlar to this line, prove that
DE=CB.
[If SY is per])endicular to the tangent at P, the triangles SYPy
CDE are similar. Then
DE:CD=SY:Sr=^S'T':S'P;
Cm-^Pr^^Cm ^^op. XXI. and Ex. 5.]
* Proposition XXXVII.
The square of the ordinate of any 'point of a hyperbola
with respect to any diameter vanes as the rectangle under
the segments of the diameter made by the ordinate.
(QV'iPV. P'V= CD'- : CP'-.)
Let QV be an ordinate to the diameter PCP', meet-
ing the asymptotes in q, q'.
Draw the tangent at P meeting the asymptotes in r, r.
Then Pr is parallel to QV. [Proj). XII.
Therefore, by similar triangles,
qT^\Pr'- = GV^:GP\
therefore q V- - Pr'- : Pr^ = G V- -GP^: GP\
I
HYPERBOLA. 171
but Pr . Pr = Qq . Qq, [Prop. XXX.
or P,^ = qV'-QV\
[Prop. XXXI. and Euc. 11. 5.
therefore qT--Pr~ = Qr\
Also Cr'-CP-' = PV.RV, [Euc. II. 5.
therefore QV- :Pi^ = PV. P'V:GP\
or QV^ : PV. FV= Pr^ : CP\
which is constant.
Since CI^ = Pr . Pr [Prop. XXX.
= Pi^, [Prop. XXXI.
this result may also be expressed as
QF2 : PV.P'V= CL^ : GPK
Ex. If the tangent at D to the conjugate hyperbola meet an
asymptote in / and the h^'perbola in q\ and the ordinate vq'
parallel to the tangent" at P be produced to meet the same asymp-
tote in R, show th&t ^CP7^=^£\CvR.
THE EQUILATERAL HYPERBOLA.
The rectangle contained by the transveree axis of a
central conic and its latus rectum has been called by
ApoUonius the " figure of the conic upon its axis." It
is evident that the "minor" or "conjugate" axis of a
central conic, according as it is an ellipse or a hyperbola,
is equal to the side of a square equivalent in area to
the "figure." (Chap. II., Prop. VI, and Chap. III..
Prop. V.)
A hyperbola which has the sides of its " figure " equal is
called an equilateral hyperbola. The latus rectum being
thus equal to the transverse axis it is clear that the con-
jugate axis is equal to the transverse axis (Chap. III.,
Prop. V.) ; in other words the two axes of an equilateral
hyperbola are equal.
172 GEOMETRY OF CONICS.
From Prop. XXVII. it is clear that the asymptotes of
an equilateral hyperbola are at right angles to each
other. From this property the curve is also called a
rectangular hyperbola.
Ex. Prove that the locus of the intersection of tangents to a
parabola including half a right angle, is a rectangular hyperbola.
(Prop. I., Ex. 10, and Prop. XXVII., Ex. 3.)
The properties of the hyperbola proved in the pre-
ceding propositions are, of course, true for the equilateral
hyperbola as well. In some cases, however, the results
assume forms which are deserving of notice.
Thus, for the equilateral hyperbola, we have
Prop. Ill, e= ^2, (See Ex. 2.)
CS=2CX.
Ex. If a circle be described on SS' as diameter, the tangents at
the vertices will intersect the asymptotes in the circumference.
Prop. V. SL = CA,
or, Latus rectum = AA'.
Prop. VIII. PN^ = AN. A'N.
Ex. 1. If PNP' be a double ordinate, the angles PAP' and
PA'P' are supplementary.
Ex. 2. The triangle formed by the tangent at any point and
its intercepts on the axes, is similar to the triangle formed by the
central radius to that point and the abscissa and ordinate of tlie
point. (See Prop. XX., Ex. 1.)
Ex. 3. If M be a point on the conjugate axis, and MP be draAvn
parallel to the transverse axis meeting the curve in P, then
PM=AM.
Ex. 4. The tangent at any point P of a circle meets a fixed
diameter AB produced in T, show that the straight line through
T perpendicular to AB meets AP BP produced in points which
lie on an equilateral hyperbola.
Ex. 5. If AB be any diameter of a circle and PNQ an ordinate
to it, the locus of intersection of jLP, BQ is an equilateral hyperbola.
HYPERBOLA. 173
Ex. 6. The locus of the point of intersection of tangents to an
ellipse which make equal angles with the majoi" and minor axis
respectively, and are not at right angles, is a rectangular hyper-
bola. (The foci of the ellipse will be tlie vertices,)
Prop. XXY[. Cir=XG,
PG = Pg = CP.
Prop. XXXL CP = Pr = Pr'.
Ex. 1. A circle whose centre is any point F and radius CP,
intei-sects the normal on the axes and the tangent on the asymp-
totes.
Ex. 2. If the tangents at two points Q and Q' meet in T, and
if CQ, CQ meet these tangents in R and R\ the circle circum-
scribing RTR passes through C
Ex. 3. The angle subtended by any chord at the centre is the
supplement of the angle between the tangents at the ends of the
chord.
Proposition A.
Coiijugate diameters are equal in the equilateral
hyperbola atid the asymptotes bisect the angle between
them.
Let CP, CD be any two conjugate semi-diameters.
Then CP" -^CD'^^CA' ^Cl^=0, [Prop. XXX^Ti.
.since the axes are equal.
Therefore CP = CD.
Again, since the asymptote Cr (Fig., Prop. XXXV.)
bisects PD it must bisect the angle PCD.
Similarly, it may be shown that the asymptote Cr
bisects the angle PCD'.
Ex. 1. A circle is described on the transverse axis as diameter.
Prove that if any tangent be di-awn to the hyperbola, the straight
lines joining tlie centre of the hyperbola with the point of contact
and with the middle point of the chord of intersection of the tan-
gent with the circle, are inclined to the asymptotes at complementary
anjiles.
174 GEOMETRY OF CONICS.
Ex. 2. The lines drawn from any point on the curve to the ex-
tremities of any diameter make equal angles with the asymptotes.
(Prop. XXXIV.)
Ex. 3. The focal chords drawn parallel to conjugate diameters
are equal. (Props. VI. and X.)
Ex. 4. If two concentric rectangular hyperbolas be described,
the axes of one being the asymptotes of the other, they will cut at
right angles.
Ex. 5. The normals at the ends of two conjugate diameters
intersect on the asymptote and are parallel to another pair of
conjugate diameters. (Prop. XXXV.)
Ex. 6. If QV be an ordinate of a diameter PCp,
QV^ = PV.p V. (Prop. XXXVIl.
Ex. 7. If tangents parallel to a given direction are drawn t(j a
system of circles passing through two fixed points, the points <>f
contact lie on a rectangular hyperbola. (Apply Ex. 6.)
Ex. 8. Given the base of a triangle and the difference of the
angles at the base, prove that the locus of the vertex is a rect-
angular hyperbola. (Apply Ex. 6.)
Ex. 9. PCp is a diameter and QF an ordinate, prove that QV in
the tangent at Q to the circle round the triangle PQp. (Ap})ly
Ex. 6.)
Ex. 10. If P be a point on an equilateral hyperbola and if tlie
tangent at Q meet CP in 7\ the circle circumscribing CTQ touches
the ordinate QV conjugate to CP. (Apply Ex. 6 and Prop. XX.)
Ex. 11. The angle between a chord PQ and the tangent at J',
is equal to the angle subtended by PQ at the other extremity of
the diameter through P.
Ex. 12. The distance of any point on the curve from the centre
is a geometric mean between its distances from the foci. (Api)lv
Prop. XXXVL, Ex. 5.)
Ex. 13. The points of intersection of an ellipse and a confocal
rectangular hyperbola are the extremities of the equi-conjugate
diameters of the ellipse. (Apply Prop. XXX VI., Ex. 5, and Chai).
II., Prop. XXXV., Ex. 5.)
Ex. 14. If two focal chords be parallel to conjugate diametei-s,
the lines joining their extremities intersect on the asymptotes.
[If P/Sj), QSq be the chords, it may be shown that pq, PQ and
an asymptote will meet on the directrix at the same point. Prop.
VII. and Prop. XXVII., Ex. 5.]
HYPEEBOLA. 175
Proposition K
In the equilateral hyperbola the transverse axis bisects.
the angle between the central radius vector of any ixtint
and the central perpeTidAcvdar on the tunaent at that
point.
Let P be any point on an equilateral hyperbola and
CD the semi-diameter conjugate to CP ; let CZ be the
perpendicular on the tangent at P.
If CR be the asymptote, because
CA = AR, [Prop. XX\ai.
the angle ACR is half a right angle, that is, half of the
angle DCZ, since CD is parallel to PZ.
[Props, XIL and XXXIII.
But the angle PCR is half of the angle PCD ; [Prop. A.
therefore the remaining angle PCA is half of the remain-
ing angle PCZ, that is, CA bisects the angle PCZ.
Ex. 1. Prove that CZ. CF== CA\ (Apply Prop. XX.)
Ex. 2. Prove that the angles CPA and CAZ are equaL
17G GEOMETRY OF CONICS.
Proposition C.
In the equilateral hyperbola diameters at inght angles
to each other are equal.
Let there be two semi-diameters GP, CD at right
angles to each other, meeting the curve and its conjugate
in P and D respectively.
Then the angle ^C5 = the angle PCA
each being a right angle. Taking away the common
angle FOB,
the angle ^CP = the angle BCD.
Hence from symmetry, since the curve and its conjugate
are equal and similarly placed with respect to the axes,
GP = CD.
Ex. 1. Prove that focal chords at right angles to each other
are equal.
Ex. 2. If a light-angled triangle be inscribed in the cui-ve, the
normal at the right angle is parallel to the hyi)otennse. (See
Prop. X.)
Ex. 3. Chords which subtend a right angle at a point P of tlie
curve, are all parallel to the normal at /'.
Proposition I).
The angle between any two diatneters of an equilateral
hyperbola is equal to the angle between their conjugates.
HYPERBOLA. 177
Let CP, CP' be any two semi-diameters, and CD, CD'
the semi-diameters conjugate to them respectively.
Then, if CR be the asymptote,
the angle PCjR = the angle DC-R, [Prop. A.
and the angle P'CjR = the angle D'CR ; [Prop. A.
therefore, by subtraction,
the angle PCP' = the angle DCD\
Ex. 1. Conjugate diameters axe inclined to either axes at angles
which are complemeutar^-.
Ex. 2. If CP, CD be conjugate semi-diameters and FX, DM
ordinates, the triangles PCS, DCM are equal in all respects.
Ex. 3. The dilFerence between the angles which the lines join-
ing any point on the curve to the extremities of a diameter make
with the diameter, is equal to the angle which the diameter makes
with its conjugate.
Ex. 4. The angles subtended by any chord at the extremities
of a diameter are equal or supplementary-. (Apply Prop. XXXIV.)
Ex. 5. AB is a chord of a circle and a diameter of a rectangvdar
hyperbola, P is any point on the circle, AP, BP, produced if
necessary, meet the hyperlx>la in Q, (Jf respectively. Prove that
BQ and A^ intersect on the circle. (Apply Ex. 4.)
Ex. 6. A circle and a rectangular hyperbola intersect in four
points and one of their common chords is a diameter of the hyper-
bola. Show that the other common chord is a diameter of the
circle. (Apply Ex. 4.)
M
178 GEOMETEY OF CONICS.
Ex. 7. QN is drawn perpendicular from any point Q on the
curve to the tangent at P. Prove that the circle round CNP
bisects PQ. (Apply Ex. 4.)
Ex. 8. If a rectangular hyperbola circumscribe a triangle, the
locus of its centre is the nine-point circle.
[The diameters to the middle points of the sides are conjugate to
the sides respectively.]
Ex. 9. The tangent at a point P of a rectangular hyperbola meets
a diameter QCQ' in T. Prove that CQ and TQ' subtend equal angles
at P.
* Proposition E.
If a rectangular hyperbola circumscribe a triangle
it passes through the orthocentre.
Let a rectangular hyperbola circumscribing a triangle
ABC meet AD, drawn perpendicular to BG, in 0.
Then the rectangles AD . OD, BD . CD are as the
squares of the semi-diameters parallel to AD, BC. [Prop. X.
But the semi-diameters being at right angles to each
other, are equal : [Prop. C.
therefore AD,OD = BD. CD.
Therefore, as is well known, the point 0 must coincide
either with the orthocentre or with the point 0' where
AD meets the circle circumscribing the triangle ABC.
i
HYPERBOLA. 179
But the latter case is impossible ; for then the lines
AD, BO, which are at right angles to each other, will be
equally inclined to the axis, [Prop. XI.
and will, therefore, be parallel to the asymptotes, which
are also at right angles to each other and equally inclined
to the axis. [Prop. XXVII.
Hence BC, being parallel t^ an a.symptote, cannot meet
the curve in two points (see Prop. XXVII., Ex. 2), which
is contrary to the hypothesis.
Hence the curve must pass through the orthocentre.
Ex. 1. Every conic passiug through the centres of the four
circles which touch the sides of a triangle is a rectangular
hv-perbola.
Ex. 2. Any conic passing through the four points of inter-
section of two rectangular hyperbolas, is itself a rectangular
h^-perbola.
Ex. 3. If two rectangular hyperbolas intersect in A, B, C\ D,
the circles described on AB, CD as diametei-s intersect each other
orthogonally.
[D is the orthocentre of the triangle ABC. Observe that the
distance between the middle jwiuts of AB and CD is equal to the
radius of the circuiuscribiug circle.]
Miscellaneous Examples on the Hyperbola.
1. Given the two asymptotes and a point on the curve,
show how to construct the curve and find the position of
the foci.
2. CP, CD are conjugate semi-diameters and the tan-
gent at P meets an asymptote in r. If m be the per-
pendicular from r on the transverse axis DPn is a right
line.
3. P is any point on a hyperbola whose foci are S, S' ;
if the tangent at P meet an asymptote in T the angle
between that asymptote and S'P is double the angle STP.
180 GEOMETEY OF CONICS.
4. Given four points on an equilateral hyperbola which
are at the extremities of two chords at right angles and
also the tangent at one of the points, find the centre of
the curve.
5. The tangents at the extremities P, P' of a chord of
a conic parallel to the transverse axis meet in T. If two
circles be drawn through Sy touching the conic at P
and P' respectively, prove that F, the second point of
intersection of the circles, will be at the intersection of
PP' and ST.
Prove also that the locus of F from different positions
of PP' will be a parabola with its vertex at S and passing
through the ends of the conjugate axis.
6. Given a pair of conjugate diameters PCP', BCD', find
the position of the axis.
[Join PD, FD', bisect them in ^and F; join CB, CF; bisect the
angle FCF by the line A'CA, and through C draw BCB' perpendi-
cular to ACA' ; these are the axes sought.]
7. If the focal radii vectores, the ordinate and the tan-
gent at any point P of a hyperbola meet an asymptote in
Q, R, E, T respectively, and M be the middle point of
QR, prove that PQ ^ PR = 2{CM ^ ET).
8. If P and Q be the points of contact of orthogonal
tangents from 0 to two confocal conies, the normals at P
and Q to the two conies will intersect on the line joining
0 to their common centre.
9. Describe the hyperbolas which have a common focus,
pass through a given point and have their asymptotes
parallel to two given straight lines.
10. From each of two points on a rectangular hyper-
bola a perpendicular is drawn on the tangent at the
HYPEEBOLA. 181
other; prove that these perpendiculars subtend equal
angles at the centre.
11. If the focal distances of a point P on a hyperbola
meet an asymptote in U and F, the perimeter of the
triangle PUV is, constant for all positions of P.
12. If a hyperbola be described touching the three sides
of a triangle, one focus lies within one of the three outer
segments of the circumscribing circle made by the sides
of the triangle.
13. Two fixed points P, Q are taken in the plane of a
given circle and a chord RS of a circle is drawn parallel
to PQ ; prove that the locus of intersection of RP and SQ
is a conic.
14. Tangents are drawn to a rectangular hyperbola from
a point T on the transverse axis, meeting the tangents at
the vertices in Q, Q. Prove that QQf touches the
auxiliary circle at R, such that RT bisects the angle
QTQ'.
15. If the tangents at the ends of a chord of a hyper-
bola meet in T and TM, TM be drawn parallel to the
asymptotes to meet them in M, M', then J/JLT is parallel
to the chord.
16. The locus of the intersection of two equal circles
which are described on two sides AB, J.C' of a triangle as
chords is a rectangular hyperbola whose centre is the
middle point of EC and which passes through A, B, C.
17. Through a fixed point 0 a chord POQ of a
hyperbola is drawn, PL, QL are drawn parallel to the
asymptotes; show that the locus of Z is a similar and
similarly situated hyperbola.
182 GEOMETRY OF CONICS.
18. A circle and a rectangular hyperbola circumscribe
a triangle ABC, right angled at C. If the tangent to the
circle at G meets the hyperbola again in C, the tangents
to the hyperbola at G, G' intersect on AB.
19. Find the locus of the middle points of a system of
chords of a hyperbola passing through a fixed point on
one of the asymptotes.
20. GP, GD are conjugate semi-diameters ; if
GD = 1J2.CB,
prove that the tangent at P passes through a focus of the
conjugate hyperbola,
21. Given a focus and three points on a conic, find the
directrix. Show that three at least of the four possible
conies must be hyperbolas.
22. The normal at any point P of a hyperbola meets
the asymptotes in g.^, g^ and the conjugate diameter in/;
prove that P/is the harmonic mean between Pg^, Pg^.
23. The sum of the squares of the perpendiculars drawn
from the foci of a hyperbola on any tangent to the conju-
gate hyperbola is constant ( = 2 . G&)
24. The tangent at P meets the asymptotes in T, t, and
the normal at P meets the transverse axis in G; prove that
the triangle TOt remains similar to itself as P varies.
25. The intercept on any tangent to a hyperbola made
by the asymptotes subtends a constant angle at either
focus.
26. Given two tangents to a rectangular hyperbola and
their points of contact, to find the asymptotes.
27. A circle touches a conic at a fixed point and cuts it
HYPERBOLA, 183
in P and Q ; the locus of the middle point of PQ is a right
line.
28. If two conies with a common directrix meet in four
points, these four points lie on a circle whose centre is on
the straight line joining the corresponding foci.
29. The locus of the middle point of a line which moves
so as to cut off a constant area from the corner of a rect-
angle is an equilateral hyperbola. (Prop. XXIX., Ex. 4.)
30. If between a rectangular hyperbola and its asymp-
totes a concentric elliptic quadrant be inscribed, the
rectangle contained by its axes is constant. (Apply
Chap. II., Prop. XXn., and Chap. III., Prop. XXIX.)
31. Given an asymptote, a tangent and its point of
contact, to construct a rectangular hyperbola.
[Let the tangent at P meet the asymptote in L. Make Pif=LP
and draw 2tIC at right angles to LC-. C is the centre and the focus
S, which lies on the bisector of the angle LCM^ is determined bv
the relation C^- = CL . CM. Prop. XXXII. The directrix bisects
C9.]
32. Straight lines, passing through a given point, are
bounded by two fixed lines at right angles to each other.
Find the locus of their middle points.
[Let OX, OF be the fixed straight lines and P the giv^en point.
If C be the middle point of OP, the locus will be a rectangular
hyperbola of which the lines through C parallel to OX and 01' are
the asymptotes. Apply Prop. XXIX.]
33. Given a point Q and a straight line AB, if a line
QCP be drawn cutting AB va. C, and P be taken in it, so
that PD being perpendicular upon AB, CD may be of
constant macmitude, the locus of P is a rectancjular
hyperbola (Prop. XXIX.)
o\. Parallel tangents are drawn to a series of confocal
184 GEOMETRY OF CONICS.
ellipses. Prove that the locus of the points of contact is a
rectangular hyperbola.
[See fisrnre, Chap. II., Prop. XXVIII. CF^CG and
PFa. PR^Ct'x. CT Therefore PF.CFo: CO. C7'=C>S2= constant.]
35. From the point of intersection of the directrix with
one of the asymptotes of a rectangular hyperbola a tangent
is drawn to the curve, meeting the other asymptote in T.
Prove that GT is equal to the transverse axis. (Apply
Prop. XXXII. and Prop. XXVIL, Ex. 5.)
36. If a rectangular hyperbola, having its asymptotes
coincident with the axes of an ellipse, touch the ellipse,
the axis of the hyperbola is a mean proportional between
the axes of the ellipse. (Apply Props. XXXL, XXXIL,
and XX.)
37. Ellipses are inscribed in a given parallelogram ;
prove that their foci lie on a rectangular hyperbola.
38. Given the centre, a tangent, and a point on a
rectangular hyperbola, find the asymptotes.
39. Prove that the parallel focal chords of conjugate
hyperbolas are to one another as the eccentricities of the
hyperbolas.
40. With each pair of three given points as foci a
hyperbola is drawn passing through the third point.
Prove that the three hyperbolas thus drawn intersect
in a point.
OLASOUW : FBINTED AT THE UNIVERSITY PBG88 BV BOBKBT MACLEHOSE.
June 1893
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